/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 41 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES (21) QDP (22) UsableRulesProof [EQUIVALENT, 0 ms] (23) QDP (24) QReductionProof [EQUIVALENT, 0 ms] (25) QDP (26) QDPSizeChangeProof [EQUIVALENT, 0 ms] (27) YES (28) QDP (29) UsableRulesProof [EQUIVALENT, 0 ms] (30) QDP (31) QReductionProof [EQUIVALENT, 0 ms] (32) QDP (33) QDPSizeChangeProof [EQUIVALENT, 0 ms] (34) YES (35) QDP (36) UsableRulesProof [EQUIVALENT, 0 ms] (37) QDP (38) QReductionProof [EQUIVALENT, 0 ms] (39) QDP (40) QDPSizeChangeProof [EQUIVALENT, 0 ms] (41) YES (42) QDP (43) UsableRulesProof [EQUIVALENT, 0 ms] (44) QDP (45) QReductionProof [EQUIVALENT, 0 ms] (46) QDP (47) QDPSizeChangeProof [EQUIVALENT, 0 ms] (48) YES (49) QDP (50) UsableRulesProof [EQUIVALENT, 0 ms] (51) QDP (52) QReductionProof [EQUIVALENT, 0 ms] (53) QDP (54) TransformationProof [EQUIVALENT, 0 ms] (55) QDP (56) TransformationProof [EQUIVALENT, 0 ms] (57) QDP (58) QDPQMonotonicMRRProof [EQUIVALENT, 129 ms] (59) QDP (60) NonInfProof [EQUIVALENT, 275 ms] (61) QDP (62) DependencyGraphProof [EQUIVALENT, 0 ms] (63) QDP (64) QDPOrderProof [EQUIVALENT, 396 ms] (65) QDP (66) NonInfProof [EQUIVALENT, 414 ms] (67) QDP (68) DependencyGraphProof [EQUIVALENT, 0 ms] (69) QDP (70) NonInfProof [EQUIVALENT, 1481 ms] (71) AND (72) QDP (73) DependencyGraphProof [EQUIVALENT, 0 ms] (74) QDP (75) QDPQMonotonicMRRProof [EQUIVALENT, 25 ms] (76) QDP (77) TransformationProof [EQUIVALENT, 0 ms] (78) QDP (79) DependencyGraphProof [EQUIVALENT, 0 ms] (80) QDP (81) TransformationProof [EQUIVALENT, 0 ms] (82) QDP (83) TransformationProof [EQUIVALENT, 0 ms] (84) QDP (85) TransformationProof [EQUIVALENT, 0 ms] (86) QDP (87) DependencyGraphProof [EQUIVALENT, 0 ms] (88) QDP (89) TransformationProof [EQUIVALENT, 0 ms] (90) QDP (91) TransformationProof [EQUIVALENT, 0 ms] (92) QDP (93) TransformationProof [EQUIVALENT, 0 ms] (94) QDP (95) TransformationProof [EQUIVALENT, 0 ms] (96) QDP (97) TransformationProof [EQUIVALENT, 0 ms] (98) QDP (99) TransformationProof [EQUIVALENT, 0 ms] (100) QDP (101) TransformationProof [EQUIVALENT, 0 ms] (102) QDP (103) TransformationProof [EQUIVALENT, 0 ms] (104) QDP (105) TransformationProof [EQUIVALENT, 0 ms] (106) QDP (107) TransformationProof [EQUIVALENT, 0 ms] (108) QDP (109) TransformationProof [EQUIVALENT, 0 ms] (110) QDP (111) UsableRulesProof [EQUIVALENT, 0 ms] (112) QDP (113) QReductionProof [EQUIVALENT, 0 ms] (114) QDP (115) TransformationProof [EQUIVALENT, 0 ms] (116) QDP (117) TransformationProof [EQUIVALENT, 0 ms] (118) QDP (119) UsableRulesProof [EQUIVALENT, 0 ms] (120) QDP (121) QReductionProof [EQUIVALENT, 0 ms] (122) QDP (123) TransformationProof [EQUIVALENT, 0 ms] (124) QDP (125) TransformationProof [EQUIVALENT, 0 ms] (126) QDP (127) TransformationProof [EQUIVALENT, 0 ms] (128) QDP (129) TransformationProof [EQUIVALENT, 0 ms] (130) QDP (131) TransformationProof [EQUIVALENT, 0 ms] (132) QDP (133) TransformationProof [EQUIVALENT, 0 ms] (134) QDP (135) TransformationProof [EQUIVALENT, 0 ms] (136) QDP (137) QDPSizeChangeProof [EQUIVALENT, 0 ms] (138) YES (139) QDP (140) DependencyGraphProof [EQUIVALENT, 0 ms] (141) QDP (142) QDPOrderProof [EQUIVALENT, 166 ms] (143) QDP (144) TransformationProof [EQUIVALENT, 0 ms] (145) QDP (146) DependencyGraphProof [EQUIVALENT, 0 ms] (147) QDP (148) TransformationProof [EQUIVALENT, 0 ms] (149) QDP (150) TransformationProof [EQUIVALENT, 0 ms] (151) QDP (152) TransformationProof [EQUIVALENT, 0 ms] (153) QDP (154) DependencyGraphProof [EQUIVALENT, 0 ms] (155) QDP (156) TransformationProof [EQUIVALENT, 0 ms] (157) QDP (158) UsableRulesProof [EQUIVALENT, 0 ms] (159) QDP (160) QReductionProof [EQUIVALENT, 0 ms] (161) QDP (162) TransformationProof [EQUIVALENT, 0 ms] (163) QDP (164) UsableRulesProof [EQUIVALENT, 0 ms] (165) QDP (166) QReductionProof [EQUIVALENT, 0 ms] (167) QDP (168) TransformationProof [EQUIVALENT, 0 ms] (169) QDP (170) TransformationProof [EQUIVALENT, 0 ms] (171) QDP (172) TransformationProof [EQUIVALENT, 0 ms] (173) QDP (174) TransformationProof [EQUIVALENT, 0 ms] (175) QDP (176) TransformationProof [EQUIVALENT, 0 ms] (177) QDP (178) QDPSizeChangeProof [EQUIVALENT, 0 ms] (179) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) empty(nil) -> true empty(cons(h, t)) -> false leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) length(nil) -> 0 length(cons(h, t)) -> s(length(t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) head(cons(h, t)) -> h tail(cons(h, t)) -> t ring(st_1, in_2, st_2, in_3, st_3, m) -> if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) if_1(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) if_2(st_1, in_2, st_2, in_3, st_3, m, true) -> if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) if_3(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) if_2(st_1, in_2, st_2, in_3, st_3, m, false) -> if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) if_4(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) if_5(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, tail(in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) if_6(st_1, in_2, st_2, in_3, st_3, m, true) -> if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) if_7(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) if_6(st_1, in_2, st_2, in_3, st_3, m, false) -> if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) if_8(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) if_9(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, in_2, st_2, tail(in_3), st_3, m) Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) length(nil) -> 0 length(cons(h, t)) -> s(length(t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) head(cons(h, t)) -> h tail(cons(h, t)) -> t fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The TRS R 2 is ring(st_1, in_2, st_2, in_3, st_3, m) -> if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) if_1(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) if_2(st_1, in_2, st_2, in_3, st_3, m, true) -> if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) if_3(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) if_2(st_1, in_2, st_2, in_3, st_3, m, false) -> if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) if_4(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) if_5(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, tail(in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) if_6(st_1, in_2, st_2, in_3, st_3, m, true) -> if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) if_7(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) if_6(st_1, in_2, st_2, in_3, st_3, m, false) -> if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) if_8(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) if_9(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, in_2, st_2, tail(in_3), st_3, m) The signature Sigma is {ring_6, if_1_7, if_2_7, if_3_7, if_4_7, if_5_7, if_6_7, if_7_7, if_8_7, if_9_7} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) empty(nil) -> true empty(cons(h, t)) -> false leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) length(nil) -> 0 length(cons(h, t)) -> s(length(t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) head(cons(h, t)) -> h tail(cons(h, t)) -> t ring(st_1, in_2, st_2, in_3, st_3, m) -> if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) if_1(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) if_2(st_1, in_2, st_2, in_3, st_3, m, true) -> if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) if_3(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) if_2(st_1, in_2, st_2, in_3, st_3, m, false) -> if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) if_4(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) if_5(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, tail(in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) if_6(st_1, in_2, st_2, in_3, st_3, m, true) -> if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) if_7(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) if_6(st_1, in_2, st_2, in_3, st_3, m, false) -> if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) if_8(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) if_9(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, in_2, st_2, tail(in_3), st_3, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: FSTSPLIT(s(n), cons(h, t)) -> FSTSPLIT(n, t) SNDSPLIT(s(n), cons(h, t)) -> SNDSPLIT(n, t) LEQ(s(n), s(m)) -> LEQ(n, m) LENGTH(cons(h, t)) -> LENGTH(t) APP(cons(h, t), x) -> APP(t, x) MAP_F(pid, cons(h, t)) -> APP(f(pid, h), map_f(pid, t)) MAP_F(pid, cons(h, t)) -> MAP_F(pid, t) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) RING(st_1, in_2, st_2, in_3, st_3, m) -> EMPTY(fstsplit(m, st_1)) RING(st_1, in_2, st_2, in_3, st_3, m) -> FSTSPLIT(m, st_1) IF_1(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) IF_1(st_1, in_2, st_2, in_3, st_3, m, false) -> SNDSPLIT(m, st_1) IF_1(st_1, in_2, st_2, in_3, st_3, m, false) -> FSTSPLIT(m, st_1) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) RING(st_1, in_2, st_2, in_3, st_3, m) -> LEQ(m, length(st_2)) RING(st_1, in_2, st_2, in_3, st_3, m) -> LENGTH(st_2) IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> EMPTY(fstsplit(m, st_2)) IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> FSTSPLIT(m, st_2) IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> SNDSPLIT(m, st_2) IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> FSTSPLIT(m, st_2) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> EMPTY(fstsplit(m, app(map_f(two, head(in_2)), st_2))) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> FSTSPLIT(m, app(map_f(two, head(in_2)), st_2)) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> APP(map_f(two, head(in_2)), st_2) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> MAP_F(two, head(in_2)) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> HEAD(in_2) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> TAIL(in_2) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> SNDSPLIT(m, app(map_f(two, head(in_2)), st_2)) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> APP(map_f(two, head(in_2)), st_2) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> MAP_F(two, head(in_2)) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> HEAD(in_2) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> FSTSPLIT(m, app(map_f(two, head(in_2)), st_2)) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) RING(st_1, in_2, st_2, in_3, st_3, m) -> EMPTY(map_f(two, head(in_2))) RING(st_1, in_2, st_2, in_3, st_3, m) -> MAP_F(two, head(in_2)) RING(st_1, in_2, st_2, in_3, st_3, m) -> HEAD(in_2) IF_5(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, tail(in_2), st_2, in_3, st_3, m) IF_5(st_1, in_2, st_2, in_3, st_3, m, true) -> TAIL(in_2) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) RING(st_1, in_2, st_2, in_3, st_3, m) -> LEQ(m, length(st_3)) RING(st_1, in_2, st_2, in_3, st_3, m) -> LENGTH(st_3) IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> EMPTY(fstsplit(m, st_3)) IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> FSTSPLIT(m, st_3) IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> SNDSPLIT(m, st_3) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> EMPTY(fstsplit(m, app(map_f(three, head(in_3)), st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> FSTSPLIT(m, app(map_f(three, head(in_3)), st_3)) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> APP(map_f(three, head(in_3)), st_3) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> MAP_F(three, head(in_3)) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> HEAD(in_3) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> TAIL(in_3) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> SNDSPLIT(m, app(map_f(three, head(in_3)), st_3)) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> APP(map_f(three, head(in_3)), st_3) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> MAP_F(three, head(in_3)) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> HEAD(in_3) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) RING(st_1, in_2, st_2, in_3, st_3, m) -> EMPTY(map_f(three, head(in_3))) RING(st_1, in_2, st_2, in_3, st_3, m) -> MAP_F(three, head(in_3)) RING(st_1, in_2, st_2, in_3, st_3, m) -> HEAD(in_3) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> TAIL(in_3) The TRS R consists of the following rules: fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) empty(nil) -> true empty(cons(h, t)) -> false leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) length(nil) -> 0 length(cons(h, t)) -> s(length(t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) head(cons(h, t)) -> h tail(cons(h, t)) -> t ring(st_1, in_2, st_2, in_3, st_3, m) -> if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) if_1(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) if_2(st_1, in_2, st_2, in_3, st_3, m, true) -> if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) if_3(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) if_2(st_1, in_2, st_2, in_3, st_3, m, false) -> if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) if_4(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) if_5(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, tail(in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) if_6(st_1, in_2, st_2, in_3, st_3, m, true) -> if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) if_7(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) if_6(st_1, in_2, st_2, in_3, st_3, m, false) -> if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) if_8(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) if_9(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, in_2, st_2, tail(in_3), st_3, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 45 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: APP(cons(h, t), x) -> APP(t, x) The TRS R consists of the following rules: fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) empty(nil) -> true empty(cons(h, t)) -> false leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) length(nil) -> 0 length(cons(h, t)) -> s(length(t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) head(cons(h, t)) -> h tail(cons(h, t)) -> t ring(st_1, in_2, st_2, in_3, st_3, m) -> if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) if_1(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) if_2(st_1, in_2, st_2, in_3, st_3, m, true) -> if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) if_3(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) if_2(st_1, in_2, st_2, in_3, st_3, m, false) -> if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) if_4(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) if_5(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, tail(in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) if_6(st_1, in_2, st_2, in_3, st_3, m, true) -> if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) if_7(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) if_6(st_1, in_2, st_2, in_3, st_3, m, false) -> if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) if_8(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) if_9(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, in_2, st_2, tail(in_3), st_3, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: APP(cons(h, t), x) -> APP(t, x) R is empty. The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: APP(cons(h, t), x) -> APP(t, x) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *APP(cons(h, t), x) -> APP(t, x) The graph contains the following edges 1 > 1, 2 >= 2 ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: MAP_F(pid, cons(h, t)) -> MAP_F(pid, t) The TRS R consists of the following rules: fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) empty(nil) -> true empty(cons(h, t)) -> false leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) length(nil) -> 0 length(cons(h, t)) -> s(length(t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) head(cons(h, t)) -> h tail(cons(h, t)) -> t ring(st_1, in_2, st_2, in_3, st_3, m) -> if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) if_1(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) if_2(st_1, in_2, st_2, in_3, st_3, m, true) -> if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) if_3(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) if_2(st_1, in_2, st_2, in_3, st_3, m, false) -> if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) if_4(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) if_5(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, tail(in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) if_6(st_1, in_2, st_2, in_3, st_3, m, true) -> if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) if_7(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) if_6(st_1, in_2, st_2, in_3, st_3, m, false) -> if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) if_8(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) if_9(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, in_2, st_2, tail(in_3), st_3, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: MAP_F(pid, cons(h, t)) -> MAP_F(pid, t) R is empty. The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: MAP_F(pid, cons(h, t)) -> MAP_F(pid, t) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *MAP_F(pid, cons(h, t)) -> MAP_F(pid, t) The graph contains the following edges 1 >= 1, 2 > 2 ---------------------------------------- (20) YES ---------------------------------------- (21) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(h, t)) -> LENGTH(t) The TRS R consists of the following rules: fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) empty(nil) -> true empty(cons(h, t)) -> false leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) length(nil) -> 0 length(cons(h, t)) -> s(length(t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) head(cons(h, t)) -> h tail(cons(h, t)) -> t ring(st_1, in_2, st_2, in_3, st_3, m) -> if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) if_1(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) if_2(st_1, in_2, st_2, in_3, st_3, m, true) -> if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) if_3(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) if_2(st_1, in_2, st_2, in_3, st_3, m, false) -> if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) if_4(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) if_5(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, tail(in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) if_6(st_1, in_2, st_2, in_3, st_3, m, true) -> if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) if_7(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) if_6(st_1, in_2, st_2, in_3, st_3, m, false) -> if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) if_8(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) if_9(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, in_2, st_2, tail(in_3), st_3, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (22) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (23) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(h, t)) -> LENGTH(t) R is empty. The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (24) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(h, t)) -> LENGTH(t) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (26) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *LENGTH(cons(h, t)) -> LENGTH(t) The graph contains the following edges 1 > 1 ---------------------------------------- (27) YES ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: LEQ(s(n), s(m)) -> LEQ(n, m) The TRS R consists of the following rules: fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) empty(nil) -> true empty(cons(h, t)) -> false leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) length(nil) -> 0 length(cons(h, t)) -> s(length(t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) head(cons(h, t)) -> h tail(cons(h, t)) -> t ring(st_1, in_2, st_2, in_3, st_3, m) -> if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) if_1(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) if_2(st_1, in_2, st_2, in_3, st_3, m, true) -> if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) if_3(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) if_2(st_1, in_2, st_2, in_3, st_3, m, false) -> if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) if_4(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) if_5(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, tail(in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) if_6(st_1, in_2, st_2, in_3, st_3, m, true) -> if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) if_7(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) if_6(st_1, in_2, st_2, in_3, st_3, m, false) -> if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) if_8(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) if_9(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, in_2, st_2, tail(in_3), st_3, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: LEQ(s(n), s(m)) -> LEQ(n, m) R is empty. The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: LEQ(s(n), s(m)) -> LEQ(n, m) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *LEQ(s(n), s(m)) -> LEQ(n, m) The graph contains the following edges 1 > 1, 2 > 2 ---------------------------------------- (34) YES ---------------------------------------- (35) Obligation: Q DP problem: The TRS P consists of the following rules: SNDSPLIT(s(n), cons(h, t)) -> SNDSPLIT(n, t) The TRS R consists of the following rules: fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) empty(nil) -> true empty(cons(h, t)) -> false leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) length(nil) -> 0 length(cons(h, t)) -> s(length(t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) head(cons(h, t)) -> h tail(cons(h, t)) -> t ring(st_1, in_2, st_2, in_3, st_3, m) -> if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) if_1(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) if_2(st_1, in_2, st_2, in_3, st_3, m, true) -> if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) if_3(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) if_2(st_1, in_2, st_2, in_3, st_3, m, false) -> if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) if_4(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) if_5(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, tail(in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) if_6(st_1, in_2, st_2, in_3, st_3, m, true) -> if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) if_7(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) if_6(st_1, in_2, st_2, in_3, st_3, m, false) -> if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) if_8(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) if_9(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, in_2, st_2, tail(in_3), st_3, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (36) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (37) Obligation: Q DP problem: The TRS P consists of the following rules: SNDSPLIT(s(n), cons(h, t)) -> SNDSPLIT(n, t) R is empty. The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (38) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) ---------------------------------------- (39) Obligation: Q DP problem: The TRS P consists of the following rules: SNDSPLIT(s(n), cons(h, t)) -> SNDSPLIT(n, t) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (40) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *SNDSPLIT(s(n), cons(h, t)) -> SNDSPLIT(n, t) The graph contains the following edges 1 > 1, 2 > 2 ---------------------------------------- (41) YES ---------------------------------------- (42) Obligation: Q DP problem: The TRS P consists of the following rules: FSTSPLIT(s(n), cons(h, t)) -> FSTSPLIT(n, t) The TRS R consists of the following rules: fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) empty(nil) -> true empty(cons(h, t)) -> false leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) length(nil) -> 0 length(cons(h, t)) -> s(length(t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) head(cons(h, t)) -> h tail(cons(h, t)) -> t ring(st_1, in_2, st_2, in_3, st_3, m) -> if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) if_1(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) if_2(st_1, in_2, st_2, in_3, st_3, m, true) -> if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) if_3(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) if_2(st_1, in_2, st_2, in_3, st_3, m, false) -> if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) if_4(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) if_5(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, tail(in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) if_6(st_1, in_2, st_2, in_3, st_3, m, true) -> if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) if_7(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) if_6(st_1, in_2, st_2, in_3, st_3, m, false) -> if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) if_8(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) if_9(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, in_2, st_2, tail(in_3), st_3, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (43) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (44) Obligation: Q DP problem: The TRS P consists of the following rules: FSTSPLIT(s(n), cons(h, t)) -> FSTSPLIT(n, t) R is empty. The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (45) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) ---------------------------------------- (46) Obligation: Q DP problem: The TRS P consists of the following rules: FSTSPLIT(s(n), cons(h, t)) -> FSTSPLIT(n, t) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (47) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *FSTSPLIT(s(n), cons(h, t)) -> FSTSPLIT(n, t) The graph contains the following edges 1 > 1, 2 > 2 ---------------------------------------- (48) YES ---------------------------------------- (49) Obligation: Q DP problem: The TRS P consists of the following rules: RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) IF_1(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) IF_5(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, tail(in_2), st_2, in_3, st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) The TRS R consists of the following rules: fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) empty(nil) -> true empty(cons(h, t)) -> false leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) length(nil) -> 0 length(cons(h, t)) -> s(length(t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) head(cons(h, t)) -> h tail(cons(h, t)) -> t ring(st_1, in_2, st_2, in_3, st_3, m) -> if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) if_1(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) if_2(st_1, in_2, st_2, in_3, st_3, m, true) -> if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) if_3(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) if_2(st_1, in_2, st_2, in_3, st_3, m, false) -> if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) if_4(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) if_5(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, tail(in_2), st_2, in_3, st_3, m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) if_6(st_1, in_2, st_2, in_3, st_3, m, true) -> if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) if_7(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) if_6(st_1, in_2, st_2, in_3, st_3, m, false) -> if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) if_8(st_1, in_2, st_2, in_3, st_3, m, false) -> ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) ring(st_1, in_2, st_2, in_3, st_3, m) -> if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) if_9(st_1, in_2, st_2, in_3, st_3, m, true) -> ring(st_1, in_2, st_2, tail(in_3), st_3, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (50) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (51) Obligation: Q DP problem: The TRS P consists of the following rules: RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) IF_1(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) IF_5(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, tail(in_2), st_2, in_3, st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (52) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. ring(x0, x1, x2, x3, x4, x5) if_1(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, true) if_3(x0, x1, x2, x3, x4, x5, false) if_2(x0, x1, x2, x3, x4, x5, false) if_4(x0, x1, x2, x3, x4, x5, false) if_5(x0, x1, x2, x3, x4, x5, true) if_6(x0, x1, x2, x3, x4, x5, true) if_7(x0, x1, x2, x3, x4, x5, false) if_6(x0, x1, x2, x3, x4, x5, false) if_8(x0, x1, x2, x3, x4, x5, false) if_9(x0, x1, x2, x3, x4, x5, true) ---------------------------------------- (53) Obligation: Q DP problem: The TRS P consists of the following rules: RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) IF_1(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) IF_5(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, tail(in_2), st_2, in_3, st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (54) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2)))) at position [6] we obtained the following new rules [LPAR04]: (RING(y0, cons(x0, x1), y2, y3, y4, y5) -> IF_5(y0, cons(x0, x1), y2, y3, y4, y5, empty(map_f(two, x0))),RING(y0, cons(x0, x1), y2, y3, y4, y5) -> IF_5(y0, cons(x0, x1), y2, y3, y4, y5, empty(map_f(two, x0)))) ---------------------------------------- (55) Obligation: Q DP problem: The TRS P consists of the following rules: RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) IF_1(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) IF_5(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, tail(in_2), st_2, in_3, st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) RING(y0, cons(x0, x1), y2, y3, y4, y5) -> IF_5(y0, cons(x0, x1), y2, y3, y4, y5, empty(map_f(two, x0))) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (56) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3)))) at position [6] we obtained the following new rules [LPAR04]: (RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))),RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0)))) ---------------------------------------- (57) Obligation: Q DP problem: The TRS P consists of the following rules: RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) IF_1(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) IF_5(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, tail(in_2), st_2, in_3, st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) RING(y0, cons(x0, x1), y2, y3, y4, y5) -> IF_5(y0, cons(x0, x1), y2, y3, y4, y5, empty(map_f(two, x0))) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (58) QDPQMonotonicMRRProof (EQUIVALENT) By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain. Strictly oriented rules of the TRS R: sndsplit(s(n), cons(h, t)) -> sndsplit(n, t) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(IF_1(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 + x_3 + x_5 POL(IF_2(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 + x_3 + x_5 POL(IF_3(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 + x_3 + x_5 POL(IF_4(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 + x_3 + x_5 POL(IF_5(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 + x_3 + x_5 POL(IF_6(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 + x_3 + x_5 POL(IF_7(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 + x_3 + x_5 POL(IF_8(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 + x_3 + x_5 POL(IF_9(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 + x_3 + x_5 POL(RING(x_1, x_2, x_3, x_4, x_5, x_6)) = x_1 + x_3 + x_5 POL(app(x_1, x_2)) = x_1 + x_2 POL(cons(x_1, x_2)) = 2 + 2*x_1 + x_2 POL(empty(x_1)) = 0 POL(f(x_1, x_2)) = 0 POL(false) = 0 POL(fstsplit(x_1, x_2)) = 2*x_2 POL(head(x_1)) = x_1 POL(length(x_1)) = 0 POL(leq(x_1, x_2)) = 0 POL(map_f(x_1, x_2)) = 0 POL(nil) = 0 POL(s(x_1)) = 0 POL(sndsplit(x_1, x_2)) = x_2 POL(tail(x_1)) = 2*x_1 POL(three) = 0 POL(true) = 0 POL(two) = 0 ---------------------------------------- (59) Obligation: Q DP problem: The TRS P consists of the following rules: RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) IF_1(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) IF_5(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, tail(in_2), st_2, in_3, st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) RING(y0, cons(x0, x1), y2, y3, y4, y5) -> IF_5(y0, cons(x0, x1), y2, y3, y4, y5, empty(map_f(two, x0))) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (60) NonInfProof (EQUIVALENT) The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: Note that final constraints are written in bold face. For Pair RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) the following chains were created: *We consider the chain IF_1(x6, x7, x8, x9, x10, x11, false) -> RING(sndsplit(x11, x6), cons(fstsplit(x11, x6), x7), x8, x9, x10, x11), RING(x12, x13, x14, x15, x16, x17) -> IF_1(x12, x13, x14, x15, x16, x17, empty(fstsplit(x17, x12))) which results in the following constraint: (1) (RING(sndsplit(x11, x6), cons(fstsplit(x11, x6), x7), x8, x9, x10, x11)=RING(x12, x13, x14, x15, x16, x17) ==> RING(x12, x13, x14, x15, x16, x17)_>=_IF_1(x12, x13, x14, x15, x16, x17, empty(fstsplit(x17, x12)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x12, x13, x8, x9, x10, x11)_>=_IF_1(x12, x13, x8, x9, x10, x11, empty(fstsplit(x11, x12)))) *We consider the chain IF_3(x30, x31, x32, x33, x34, x35, false) -> RING(x30, x31, sndsplit(x35, x32), cons(fstsplit(x35, x32), x33), x34, x35), RING(x36, x37, x38, x39, x40, x41) -> IF_1(x36, x37, x38, x39, x40, x41, empty(fstsplit(x41, x36))) which results in the following constraint: (1) (RING(x30, x31, sndsplit(x35, x32), cons(fstsplit(x35, x32), x33), x34, x35)=RING(x36, x37, x38, x39, x40, x41) ==> RING(x36, x37, x38, x39, x40, x41)_>=_IF_1(x36, x37, x38, x39, x40, x41, empty(fstsplit(x41, x36)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x30, x31, x38, x39, x34, x35)_>=_IF_1(x30, x31, x38, x39, x34, x35, empty(fstsplit(x35, x30)))) *We consider the chain IF_5(x42, x43, x44, x45, x46, x47, true) -> RING(x42, tail(x43), x44, x45, x46, x47), RING(x48, x49, x50, x51, x52, x53) -> IF_1(x48, x49, x50, x51, x52, x53, empty(fstsplit(x53, x48))) which results in the following constraint: (1) (RING(x42, tail(x43), x44, x45, x46, x47)=RING(x48, x49, x50, x51, x52, x53) ==> RING(x48, x49, x50, x51, x52, x53)_>=_IF_1(x48, x49, x50, x51, x52, x53, empty(fstsplit(x53, x48)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x42, x49, x44, x45, x46, x47)_>=_IF_1(x42, x49, x44, x45, x46, x47, empty(fstsplit(x47, x42)))) *We consider the chain IF_7(x66, x67, x68, x69, x70, x71, false) -> RING(x66, x67, x68, x69, sndsplit(x71, x70), x71), RING(x72, x73, x74, x75, x76, x77) -> IF_1(x72, x73, x74, x75, x76, x77, empty(fstsplit(x77, x72))) which results in the following constraint: (1) (RING(x66, x67, x68, x69, sndsplit(x71, x70), x71)=RING(x72, x73, x74, x75, x76, x77) ==> RING(x72, x73, x74, x75, x76, x77)_>=_IF_1(x72, x73, x74, x75, x76, x77, empty(fstsplit(x77, x72)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x66, x67, x68, x69, x76, x71)_>=_IF_1(x66, x67, x68, x69, x76, x71, empty(fstsplit(x71, x66)))) *We consider the chain IF_9(x78, x79, x80, x81, x82, x83, true) -> RING(x78, x79, x80, tail(x81), x82, x83), RING(x84, x85, x86, x87, x88, x89) -> IF_1(x84, x85, x86, x87, x88, x89, empty(fstsplit(x89, x84))) which results in the following constraint: (1) (RING(x78, x79, x80, tail(x81), x82, x83)=RING(x84, x85, x86, x87, x88, x89) ==> RING(x84, x85, x86, x87, x88, x89)_>=_IF_1(x84, x85, x86, x87, x88, x89, empty(fstsplit(x89, x84)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x78, x79, x80, x87, x82, x83)_>=_IF_1(x78, x79, x80, x87, x82, x83, empty(fstsplit(x83, x78)))) *We consider the chain IF_8(x96, x97, x98, x99, x100, x101, false) -> RING(x96, x97, x98, tail(x99), sndsplit(x101, app(map_f(three, head(x99)), x100)), x101), RING(x102, x103, x104, x105, x106, x107) -> IF_1(x102, x103, x104, x105, x106, x107, empty(fstsplit(x107, x102))) which results in the following constraint: (1) (RING(x96, x97, x98, tail(x99), sndsplit(x101, app(map_f(three, head(x99)), x100)), x101)=RING(x102, x103, x104, x105, x106, x107) ==> RING(x102, x103, x104, x105, x106, x107)_>=_IF_1(x102, x103, x104, x105, x106, x107, empty(fstsplit(x107, x102)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x96, x97, x98, x105, x106, x101)_>=_IF_1(x96, x97, x98, x105, x106, x101, empty(fstsplit(x101, x96)))) *We consider the chain IF_4(x114, x115, x116, x117, x118, x119, false) -> RING(x114, tail(x115), sndsplit(x119, app(map_f(two, head(x115)), x116)), cons(fstsplit(x119, app(map_f(two, head(x115)), x116)), x117), x118, x119), RING(x120, x121, x122, x123, x124, x125) -> IF_1(x120, x121, x122, x123, x124, x125, empty(fstsplit(x125, x120))) which results in the following constraint: (1) (RING(x114, tail(x115), sndsplit(x119, app(map_f(two, head(x115)), x116)), cons(fstsplit(x119, app(map_f(two, head(x115)), x116)), x117), x118, x119)=RING(x120, x121, x122, x123, x124, x125) ==> RING(x120, x121, x122, x123, x124, x125)_>=_IF_1(x120, x121, x122, x123, x124, x125, empty(fstsplit(x125, x120)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x114, x121, x122, x123, x118, x119)_>=_IF_1(x114, x121, x122, x123, x118, x119, empty(fstsplit(x119, x114)))) For Pair IF_1(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) the following chains were created: *We consider the chain RING(x140, x141, x142, x143, x144, x145) -> IF_1(x140, x141, x142, x143, x144, x145, empty(fstsplit(x145, x140))), IF_1(x146, x147, x148, x149, x150, x151, false) -> RING(sndsplit(x151, x146), cons(fstsplit(x151, x146), x147), x148, x149, x150, x151) which results in the following constraint: (1) (IF_1(x140, x141, x142, x143, x144, x145, empty(fstsplit(x145, x140)))=IF_1(x146, x147, x148, x149, x150, x151, false) ==> IF_1(x146, x147, x148, x149, x150, x151, false)_>=_RING(sndsplit(x151, x146), cons(fstsplit(x151, x146), x147), x148, x149, x150, x151)) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (fstsplit(x145, x140)=x1866 & empty(x1866)=false ==> IF_1(x140, x141, x142, x143, x144, x145, false)_>=_RING(sndsplit(x145, x140), cons(fstsplit(x145, x140), x141), x142, x143, x144, x145)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x1866)=false which results in the following new constraint: (3) (false=false & fstsplit(x145, x140)=cons(x1868, x1867) ==> IF_1(x140, x141, x142, x143, x144, x145, false)_>=_RING(sndsplit(x145, x140), cons(fstsplit(x145, x140), x141), x142, x143, x144, x145)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (fstsplit(x145, x140)=cons(x1868, x1867) ==> IF_1(x140, x141, x142, x143, x144, x145, false)_>=_RING(sndsplit(x145, x140), cons(fstsplit(x145, x140), x141), x142, x143, x144, x145)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on fstsplit(x145, x140)=cons(x1868, x1867) which results in the following new constraint: (5) (cons(x1872, fstsplit(x1873, x1871))=cons(x1868, x1867) & (\/x1874,x1875,x1876,x1877,x1878,x1879:fstsplit(x1873, x1871)=cons(x1874, x1875) ==> IF_1(x1871, x1876, x1877, x1878, x1879, x1873, false)_>=_RING(sndsplit(x1873, x1871), cons(fstsplit(x1873, x1871), x1876), x1877, x1878, x1879, x1873)) ==> IF_1(cons(x1872, x1871), x141, x142, x143, x144, s(x1873), false)_>=_RING(sndsplit(s(x1873), cons(x1872, x1871)), cons(fstsplit(s(x1873), cons(x1872, x1871)), x141), x142, x143, x144, s(x1873))) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (6) (IF_1(cons(x1872, x1871), x141, x142, x143, x144, s(x1873), false)_>=_RING(sndsplit(s(x1873), cons(x1872, x1871)), cons(fstsplit(s(x1873), cons(x1872, x1871)), x141), x142, x143, x144, s(x1873))) For Pair RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) the following chains were created: *We consider the chain IF_1(x250, x251, x252, x253, x254, x255, false) -> RING(sndsplit(x255, x250), cons(fstsplit(x255, x250), x251), x252, x253, x254, x255), RING(x256, x257, x258, x259, x260, x261) -> IF_2(x256, x257, x258, x259, x260, x261, leq(x261, length(x258))) which results in the following constraint: (1) (RING(sndsplit(x255, x250), cons(fstsplit(x255, x250), x251), x252, x253, x254, x255)=RING(x256, x257, x258, x259, x260, x261) ==> RING(x256, x257, x258, x259, x260, x261)_>=_IF_2(x256, x257, x258, x259, x260, x261, leq(x261, length(x258)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x256, x257, x252, x253, x254, x255)_>=_IF_2(x256, x257, x252, x253, x254, x255, leq(x255, length(x252)))) *We consider the chain IF_3(x274, x275, x276, x277, x278, x279, false) -> RING(x274, x275, sndsplit(x279, x276), cons(fstsplit(x279, x276), x277), x278, x279), RING(x280, x281, x282, x283, x284, x285) -> IF_2(x280, x281, x282, x283, x284, x285, leq(x285, length(x282))) which results in the following constraint: (1) (RING(x274, x275, sndsplit(x279, x276), cons(fstsplit(x279, x276), x277), x278, x279)=RING(x280, x281, x282, x283, x284, x285) ==> RING(x280, x281, x282, x283, x284, x285)_>=_IF_2(x280, x281, x282, x283, x284, x285, leq(x285, length(x282)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x274, x275, x282, x283, x278, x279)_>=_IF_2(x274, x275, x282, x283, x278, x279, leq(x279, length(x282)))) *We consider the chain IF_5(x286, x287, x288, x289, x290, x291, true) -> RING(x286, tail(x287), x288, x289, x290, x291), RING(x292, x293, x294, x295, x296, x297) -> IF_2(x292, x293, x294, x295, x296, x297, leq(x297, length(x294))) which results in the following constraint: (1) (RING(x286, tail(x287), x288, x289, x290, x291)=RING(x292, x293, x294, x295, x296, x297) ==> RING(x292, x293, x294, x295, x296, x297)_>=_IF_2(x292, x293, x294, x295, x296, x297, leq(x297, length(x294)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x286, x293, x288, x289, x290, x291)_>=_IF_2(x286, x293, x288, x289, x290, x291, leq(x291, length(x288)))) *We consider the chain IF_7(x310, x311, x312, x313, x314, x315, false) -> RING(x310, x311, x312, x313, sndsplit(x315, x314), x315), RING(x316, x317, x318, x319, x320, x321) -> IF_2(x316, x317, x318, x319, x320, x321, leq(x321, length(x318))) which results in the following constraint: (1) (RING(x310, x311, x312, x313, sndsplit(x315, x314), x315)=RING(x316, x317, x318, x319, x320, x321) ==> RING(x316, x317, x318, x319, x320, x321)_>=_IF_2(x316, x317, x318, x319, x320, x321, leq(x321, length(x318)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x310, x311, x312, x313, x320, x315)_>=_IF_2(x310, x311, x312, x313, x320, x315, leq(x315, length(x312)))) *We consider the chain IF_9(x322, x323, x324, x325, x326, x327, true) -> RING(x322, x323, x324, tail(x325), x326, x327), RING(x328, x329, x330, x331, x332, x333) -> IF_2(x328, x329, x330, x331, x332, x333, leq(x333, length(x330))) which results in the following constraint: (1) (RING(x322, x323, x324, tail(x325), x326, x327)=RING(x328, x329, x330, x331, x332, x333) ==> RING(x328, x329, x330, x331, x332, x333)_>=_IF_2(x328, x329, x330, x331, x332, x333, leq(x333, length(x330)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x322, x323, x324, x331, x326, x327)_>=_IF_2(x322, x323, x324, x331, x326, x327, leq(x327, length(x324)))) *We consider the chain IF_8(x340, x341, x342, x343, x344, x345, false) -> RING(x340, x341, x342, tail(x343), sndsplit(x345, app(map_f(three, head(x343)), x344)), x345), RING(x346, x347, x348, x349, x350, x351) -> IF_2(x346, x347, x348, x349, x350, x351, leq(x351, length(x348))) which results in the following constraint: (1) (RING(x340, x341, x342, tail(x343), sndsplit(x345, app(map_f(three, head(x343)), x344)), x345)=RING(x346, x347, x348, x349, x350, x351) ==> RING(x346, x347, x348, x349, x350, x351)_>=_IF_2(x346, x347, x348, x349, x350, x351, leq(x351, length(x348)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x340, x341, x342, x349, x350, x345)_>=_IF_2(x340, x341, x342, x349, x350, x345, leq(x345, length(x342)))) *We consider the chain IF_4(x358, x359, x360, x361, x362, x363, false) -> RING(x358, tail(x359), sndsplit(x363, app(map_f(two, head(x359)), x360)), cons(fstsplit(x363, app(map_f(two, head(x359)), x360)), x361), x362, x363), RING(x364, x365, x366, x367, x368, x369) -> IF_2(x364, x365, x366, x367, x368, x369, leq(x369, length(x366))) which results in the following constraint: (1) (RING(x358, tail(x359), sndsplit(x363, app(map_f(two, head(x359)), x360)), cons(fstsplit(x363, app(map_f(two, head(x359)), x360)), x361), x362, x363)=RING(x364, x365, x366, x367, x368, x369) ==> RING(x364, x365, x366, x367, x368, x369)_>=_IF_2(x364, x365, x366, x367, x368, x369, leq(x369, length(x366)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x358, x365, x366, x367, x362, x363)_>=_IF_2(x358, x365, x366, x367, x362, x363, leq(x363, length(x366)))) For Pair IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) the following chains were created: *We consider the chain RING(x396, x397, x398, x399, x400, x401) -> IF_2(x396, x397, x398, x399, x400, x401, leq(x401, length(x398))), IF_2(x402, x403, x404, x405, x406, x407, true) -> IF_3(x402, x403, x404, x405, x406, x407, empty(fstsplit(x407, x404))) which results in the following constraint: (1) (IF_2(x396, x397, x398, x399, x400, x401, leq(x401, length(x398)))=IF_2(x402, x403, x404, x405, x406, x407, true) ==> IF_2(x402, x403, x404, x405, x406, x407, true)_>=_IF_3(x402, x403, x404, x405, x406, x407, empty(fstsplit(x407, x404)))) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (length(x398)=x1888 & leq(x401, x1888)=true ==> IF_2(x396, x397, x398, x399, x400, x401, true)_>=_IF_3(x396, x397, x398, x399, x400, x401, empty(fstsplit(x401, x398)))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on leq(x401, x1888)=true which results in the following new constraints: (3) (true=true & length(x398)=x1889 ==> IF_2(x396, x397, x398, x399, x400, 0, true)_>=_IF_3(x396, x397, x398, x399, x400, 0, empty(fstsplit(0, x398)))) (4) (leq(x1892, x1891)=true & length(x398)=s(x1891) & (\/x1893,x1894,x1895,x1896,x1897:leq(x1892, x1891)=true & length(x1893)=x1891 ==> IF_2(x1894, x1895, x1893, x1896, x1897, x1892, true)_>=_IF_3(x1894, x1895, x1893, x1896, x1897, x1892, empty(fstsplit(x1892, x1893)))) ==> IF_2(x396, x397, x398, x399, x400, s(x1892), true)_>=_IF_3(x396, x397, x398, x399, x400, s(x1892), empty(fstsplit(s(x1892), x398)))) We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint: (5) (IF_2(x396, x397, x398, x399, x400, 0, true)_>=_IF_3(x396, x397, x398, x399, x400, 0, empty(fstsplit(0, x398)))) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on length(x398)=s(x1891) which results in the following new constraint: (6) (s(length(x1898))=s(x1891) & leq(x1892, x1891)=true & (\/x1893,x1894,x1895,x1896,x1897:leq(x1892, x1891)=true & length(x1893)=x1891 ==> IF_2(x1894, x1895, x1893, x1896, x1897, x1892, true)_>=_IF_3(x1894, x1895, x1893, x1896, x1897, x1892, empty(fstsplit(x1892, x1893)))) & (\/x1900,x1901,x1902,x1903,x1904,x1905,x1906,x1907,x1908,x1909,x1910:length(x1898)=s(x1900) & leq(x1901, x1900)=true & (\/x1902,x1903,x1904,x1905,x1906:leq(x1901, x1900)=true & length(x1902)=x1900 ==> IF_2(x1903, x1904, x1902, x1905, x1906, x1901, true)_>=_IF_3(x1903, x1904, x1902, x1905, x1906, x1901, empty(fstsplit(x1901, x1902)))) ==> IF_2(x1907, x1908, x1898, x1909, x1910, s(x1901), true)_>=_IF_3(x1907, x1908, x1898, x1909, x1910, s(x1901), empty(fstsplit(s(x1901), x1898)))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) We simplified constraint (6) using rules (I), (II) which results in the following new constraint: (7) (length(x1898)=x1891 & leq(x1892, x1891)=true & (\/x1893,x1894,x1895,x1896,x1897:leq(x1892, x1891)=true & length(x1893)=x1891 ==> IF_2(x1894, x1895, x1893, x1896, x1897, x1892, true)_>=_IF_3(x1894, x1895, x1893, x1896, x1897, x1892, empty(fstsplit(x1892, x1893)))) & (\/x1900,x1901,x1902,x1903,x1904,x1905,x1906,x1907,x1908,x1909,x1910:length(x1898)=s(x1900) & leq(x1901, x1900)=true & (\/x1902,x1903,x1904,x1905,x1906:leq(x1901, x1900)=true & length(x1902)=x1900 ==> IF_2(x1903, x1904, x1902, x1905, x1906, x1901, true)_>=_IF_3(x1903, x1904, x1902, x1905, x1906, x1901, empty(fstsplit(x1901, x1902)))) ==> IF_2(x1907, x1908, x1898, x1909, x1910, s(x1901), true)_>=_IF_3(x1907, x1908, x1898, x1909, x1910, s(x1901), empty(fstsplit(s(x1901), x1898)))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) We simplified constraint (7) using rule (VI) where we applied the induction hypothesis (\/x1893,x1894,x1895,x1896,x1897:leq(x1892, x1891)=true & length(x1893)=x1891 ==> IF_2(x1894, x1895, x1893, x1896, x1897, x1892, true)_>=_IF_3(x1894, x1895, x1893, x1896, x1897, x1892, empty(fstsplit(x1892, x1893)))) with sigma = [x1893 / x1898, x1894 / x396, x1895 / x397, x1896 / x399, x1897 / x400] which results in the following new constraint: (8) (IF_2(x396, x397, x1898, x399, x400, x1892, true)_>=_IF_3(x396, x397, x1898, x399, x400, x1892, empty(fstsplit(x1892, x1898))) & (\/x1900,x1901,x1902,x1903,x1904,x1905,x1906,x1907,x1908,x1909,x1910:length(x1898)=s(x1900) & leq(x1901, x1900)=true & (\/x1902,x1903,x1904,x1905,x1906:leq(x1901, x1900)=true & length(x1902)=x1900 ==> IF_2(x1903, x1904, x1902, x1905, x1906, x1901, true)_>=_IF_3(x1903, x1904, x1902, x1905, x1906, x1901, empty(fstsplit(x1901, x1902)))) ==> IF_2(x1907, x1908, x1898, x1909, x1910, s(x1901), true)_>=_IF_3(x1907, x1908, x1898, x1909, x1910, s(x1901), empty(fstsplit(s(x1901), x1898)))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) We simplified constraint (8) using rule (IV) which results in the following new constraint: (9) (IF_2(x396, x397, x1898, x399, x400, x1892, true)_>=_IF_3(x396, x397, x1898, x399, x400, x1892, empty(fstsplit(x1892, x1898))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) For Pair IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) the following chains were created: *We consider the chain IF_2(x506, x507, x508, x509, x510, x511, true) -> IF_3(x506, x507, x508, x509, x510, x511, empty(fstsplit(x511, x508))), IF_3(x512, x513, x514, x515, x516, x517, false) -> RING(x512, x513, sndsplit(x517, x514), cons(fstsplit(x517, x514), x515), x516, x517) which results in the following constraint: (1) (IF_3(x506, x507, x508, x509, x510, x511, empty(fstsplit(x511, x508)))=IF_3(x512, x513, x514, x515, x516, x517, false) ==> IF_3(x512, x513, x514, x515, x516, x517, false)_>=_RING(x512, x513, sndsplit(x517, x514), cons(fstsplit(x517, x514), x515), x516, x517)) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (fstsplit(x511, x508)=x1911 & empty(x1911)=false ==> IF_3(x506, x507, x508, x509, x510, x511, false)_>=_RING(x506, x507, sndsplit(x511, x508), cons(fstsplit(x511, x508), x509), x510, x511)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x1911)=false which results in the following new constraint: (3) (false=false & fstsplit(x511, x508)=cons(x1913, x1912) ==> IF_3(x506, x507, x508, x509, x510, x511, false)_>=_RING(x506, x507, sndsplit(x511, x508), cons(fstsplit(x511, x508), x509), x510, x511)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (fstsplit(x511, x508)=cons(x1913, x1912) ==> IF_3(x506, x507, x508, x509, x510, x511, false)_>=_RING(x506, x507, sndsplit(x511, x508), cons(fstsplit(x511, x508), x509), x510, x511)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on fstsplit(x511, x508)=cons(x1913, x1912) which results in the following new constraint: (5) (cons(x1917, fstsplit(x1918, x1916))=cons(x1913, x1912) & (\/x1919,x1920,x1921,x1922,x1923,x1924:fstsplit(x1918, x1916)=cons(x1919, x1920) ==> IF_3(x1921, x1922, x1916, x1923, x1924, x1918, false)_>=_RING(x1921, x1922, sndsplit(x1918, x1916), cons(fstsplit(x1918, x1916), x1923), x1924, x1918)) ==> IF_3(x506, x507, cons(x1917, x1916), x509, x510, s(x1918), false)_>=_RING(x506, x507, sndsplit(s(x1918), cons(x1917, x1916)), cons(fstsplit(s(x1918), cons(x1917, x1916)), x509), x510, s(x1918))) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (6) (IF_3(x506, x507, cons(x1917, x1916), x509, x510, s(x1918), false)_>=_RING(x506, x507, sndsplit(s(x1918), cons(x1917, x1916)), cons(fstsplit(s(x1918), cons(x1917, x1916)), x509), x510, s(x1918))) For Pair IF_5(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, tail(in_2), st_2, in_3, st_3, m) the following chains were created: *We consider the chain RING(x676, cons(x677, x678), x679, x680, x681, x682) -> IF_5(x676, cons(x677, x678), x679, x680, x681, x682, empty(map_f(two, x677))), IF_5(x683, x684, x685, x686, x687, x688, true) -> RING(x683, tail(x684), x685, x686, x687, x688) which results in the following constraint: (1) (IF_5(x676, cons(x677, x678), x679, x680, x681, x682, empty(map_f(two, x677)))=IF_5(x683, x684, x685, x686, x687, x688, true) ==> IF_5(x683, x684, x685, x686, x687, x688, true)_>=_RING(x683, tail(x684), x685, x686, x687, x688)) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (two=x1926 & map_f(x1926, x677)=x1925 & empty(x1925)=true ==> IF_5(x676, cons(x677, x678), x679, x680, x681, x682, true)_>=_RING(x676, tail(cons(x677, x678)), x679, x680, x681, x682)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x1925)=true which results in the following new constraint: (3) (true=true & two=x1926 & map_f(x1926, x677)=nil ==> IF_5(x676, cons(x677, x678), x679, x680, x681, x682, true)_>=_RING(x676, tail(cons(x677, x678)), x679, x680, x681, x682)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (two=x1926 & map_f(x1926, x677)=nil ==> IF_5(x676, cons(x677, x678), x679, x680, x681, x682, true)_>=_RING(x676, tail(cons(x677, x678)), x679, x680, x681, x682)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on map_f(x1926, x677)=nil which results in the following new constraints: (5) (nil=nil & two=x1929 ==> IF_5(x676, cons(nil, x678), x679, x680, x681, x682, true)_>=_RING(x676, tail(cons(nil, x678)), x679, x680, x681, x682)) (6) (app(f(x1932, x1931), map_f(x1932, x1930))=nil & two=x1932 & (\/x1933,x1934,x1935,x1936,x1937,x1938:map_f(x1932, x1930)=nil & two=x1932 ==> IF_5(x1933, cons(x1930, x1934), x1935, x1936, x1937, x1938, true)_>=_RING(x1933, tail(cons(x1930, x1934)), x1935, x1936, x1937, x1938)) ==> IF_5(x676, cons(cons(x1931, x1930), x678), x679, x680, x681, x682, true)_>=_RING(x676, tail(cons(cons(x1931, x1930), x678)), x679, x680, x681, x682)) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (7) (IF_5(x676, cons(nil, x678), x679, x680, x681, x682, true)_>=_RING(x676, tail(cons(nil, x678)), x679, x680, x681, x682)) We simplified constraint (6) using rules (IV), (VII) which results in the following new constraint: (8) (f(x1932, x1931)=x1939 & map_f(x1932, x1930)=x1940 & app(x1939, x1940)=nil & two=x1932 ==> IF_5(x676, cons(cons(x1931, x1930), x678), x679, x680, x681, x682, true)_>=_RING(x676, tail(cons(cons(x1931, x1930), x678)), x679, x680, x681, x682)) We simplified constraint (8) using rule (V) (with possible (I) afterwards) using induction on app(x1939, x1940)=nil which results in the following new constraint: (9) (x1941=nil & f(x1932, x1931)=nil & map_f(x1932, x1930)=x1941 & two=x1932 ==> IF_5(x676, cons(cons(x1931, x1930), x678), x679, x680, x681, x682, true)_>=_RING(x676, tail(cons(cons(x1931, x1930), x678)), x679, x680, x681, x682)) We solved constraint (9) using rules (I), (II). For Pair RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) the following chains were created: *We consider the chain IF_1(x702, x703, x704, x705, x706, x707, false) -> RING(sndsplit(x707, x702), cons(fstsplit(x707, x702), x703), x704, x705, x706, x707), RING(x708, x709, x710, x711, x712, x713) -> IF_6(x708, x709, x710, x711, x712, x713, leq(x713, length(x712))) which results in the following constraint: (1) (RING(sndsplit(x707, x702), cons(fstsplit(x707, x702), x703), x704, x705, x706, x707)=RING(x708, x709, x710, x711, x712, x713) ==> RING(x708, x709, x710, x711, x712, x713)_>=_IF_6(x708, x709, x710, x711, x712, x713, leq(x713, length(x712)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x708, x709, x704, x705, x706, x707)_>=_IF_6(x708, x709, x704, x705, x706, x707, leq(x707, length(x706)))) *We consider the chain IF_3(x726, x727, x728, x729, x730, x731, false) -> RING(x726, x727, sndsplit(x731, x728), cons(fstsplit(x731, x728), x729), x730, x731), RING(x732, x733, x734, x735, x736, x737) -> IF_6(x732, x733, x734, x735, x736, x737, leq(x737, length(x736))) which results in the following constraint: (1) (RING(x726, x727, sndsplit(x731, x728), cons(fstsplit(x731, x728), x729), x730, x731)=RING(x732, x733, x734, x735, x736, x737) ==> RING(x732, x733, x734, x735, x736, x737)_>=_IF_6(x732, x733, x734, x735, x736, x737, leq(x737, length(x736)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x726, x727, x734, x735, x730, x731)_>=_IF_6(x726, x727, x734, x735, x730, x731, leq(x731, length(x730)))) *We consider the chain IF_5(x738, x739, x740, x741, x742, x743, true) -> RING(x738, tail(x739), x740, x741, x742, x743), RING(x744, x745, x746, x747, x748, x749) -> IF_6(x744, x745, x746, x747, x748, x749, leq(x749, length(x748))) which results in the following constraint: (1) (RING(x738, tail(x739), x740, x741, x742, x743)=RING(x744, x745, x746, x747, x748, x749) ==> RING(x744, x745, x746, x747, x748, x749)_>=_IF_6(x744, x745, x746, x747, x748, x749, leq(x749, length(x748)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x738, x745, x740, x741, x742, x743)_>=_IF_6(x738, x745, x740, x741, x742, x743, leq(x743, length(x742)))) *We consider the chain IF_7(x762, x763, x764, x765, x766, x767, false) -> RING(x762, x763, x764, x765, sndsplit(x767, x766), x767), RING(x768, x769, x770, x771, x772, x773) -> IF_6(x768, x769, x770, x771, x772, x773, leq(x773, length(x772))) which results in the following constraint: (1) (RING(x762, x763, x764, x765, sndsplit(x767, x766), x767)=RING(x768, x769, x770, x771, x772, x773) ==> RING(x768, x769, x770, x771, x772, x773)_>=_IF_6(x768, x769, x770, x771, x772, x773, leq(x773, length(x772)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x762, x763, x764, x765, x772, x767)_>=_IF_6(x762, x763, x764, x765, x772, x767, leq(x767, length(x772)))) *We consider the chain IF_9(x774, x775, x776, x777, x778, x779, true) -> RING(x774, x775, x776, tail(x777), x778, x779), RING(x780, x781, x782, x783, x784, x785) -> IF_6(x780, x781, x782, x783, x784, x785, leq(x785, length(x784))) which results in the following constraint: (1) (RING(x774, x775, x776, tail(x777), x778, x779)=RING(x780, x781, x782, x783, x784, x785) ==> RING(x780, x781, x782, x783, x784, x785)_>=_IF_6(x780, x781, x782, x783, x784, x785, leq(x785, length(x784)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x774, x775, x776, x783, x778, x779)_>=_IF_6(x774, x775, x776, x783, x778, x779, leq(x779, length(x778)))) *We consider the chain IF_8(x792, x793, x794, x795, x796, x797, false) -> RING(x792, x793, x794, tail(x795), sndsplit(x797, app(map_f(three, head(x795)), x796)), x797), RING(x798, x799, x800, x801, x802, x803) -> IF_6(x798, x799, x800, x801, x802, x803, leq(x803, length(x802))) which results in the following constraint: (1) (RING(x792, x793, x794, tail(x795), sndsplit(x797, app(map_f(three, head(x795)), x796)), x797)=RING(x798, x799, x800, x801, x802, x803) ==> RING(x798, x799, x800, x801, x802, x803)_>=_IF_6(x798, x799, x800, x801, x802, x803, leq(x803, length(x802)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x792, x793, x794, x801, x802, x797)_>=_IF_6(x792, x793, x794, x801, x802, x797, leq(x797, length(x802)))) *We consider the chain IF_4(x810, x811, x812, x813, x814, x815, false) -> RING(x810, tail(x811), sndsplit(x815, app(map_f(two, head(x811)), x812)), cons(fstsplit(x815, app(map_f(two, head(x811)), x812)), x813), x814, x815), RING(x816, x817, x818, x819, x820, x821) -> IF_6(x816, x817, x818, x819, x820, x821, leq(x821, length(x820))) which results in the following constraint: (1) (RING(x810, tail(x811), sndsplit(x815, app(map_f(two, head(x811)), x812)), cons(fstsplit(x815, app(map_f(two, head(x811)), x812)), x813), x814, x815)=RING(x816, x817, x818, x819, x820, x821) ==> RING(x816, x817, x818, x819, x820, x821)_>=_IF_6(x816, x817, x818, x819, x820, x821, leq(x821, length(x820)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x810, x817, x818, x819, x814, x815)_>=_IF_6(x810, x817, x818, x819, x814, x815, leq(x815, length(x814)))) For Pair IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) the following chains were created: *We consider the chain RING(x872, x873, x874, x875, x876, x877) -> IF_6(x872, x873, x874, x875, x876, x877, leq(x877, length(x876))), IF_6(x878, x879, x880, x881, x882, x883, true) -> IF_7(x878, x879, x880, x881, x882, x883, empty(fstsplit(x883, x882))) which results in the following constraint: (1) (IF_6(x872, x873, x874, x875, x876, x877, leq(x877, length(x876)))=IF_6(x878, x879, x880, x881, x882, x883, true) ==> IF_6(x878, x879, x880, x881, x882, x883, true)_>=_IF_7(x878, x879, x880, x881, x882, x883, empty(fstsplit(x883, x882)))) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (length(x876)=x1953 & leq(x877, x1953)=true ==> IF_6(x872, x873, x874, x875, x876, x877, true)_>=_IF_7(x872, x873, x874, x875, x876, x877, empty(fstsplit(x877, x876)))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on leq(x877, x1953)=true which results in the following new constraints: (3) (true=true & length(x876)=x1954 ==> IF_6(x872, x873, x874, x875, x876, 0, true)_>=_IF_7(x872, x873, x874, x875, x876, 0, empty(fstsplit(0, x876)))) (4) (leq(x1957, x1956)=true & length(x876)=s(x1956) & (\/x1958,x1959,x1960,x1961,x1962:leq(x1957, x1956)=true & length(x1958)=x1956 ==> IF_6(x1959, x1960, x1961, x1962, x1958, x1957, true)_>=_IF_7(x1959, x1960, x1961, x1962, x1958, x1957, empty(fstsplit(x1957, x1958)))) ==> IF_6(x872, x873, x874, x875, x876, s(x1957), true)_>=_IF_7(x872, x873, x874, x875, x876, s(x1957), empty(fstsplit(s(x1957), x876)))) We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint: (5) (IF_6(x872, x873, x874, x875, x876, 0, true)_>=_IF_7(x872, x873, x874, x875, x876, 0, empty(fstsplit(0, x876)))) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on length(x876)=s(x1956) which results in the following new constraint: (6) (s(length(x1963))=s(x1956) & leq(x1957, x1956)=true & (\/x1958,x1959,x1960,x1961,x1962:leq(x1957, x1956)=true & length(x1958)=x1956 ==> IF_6(x1959, x1960, x1961, x1962, x1958, x1957, true)_>=_IF_7(x1959, x1960, x1961, x1962, x1958, x1957, empty(fstsplit(x1957, x1958)))) & (\/x1965,x1966,x1967,x1968,x1969,x1970,x1971,x1972,x1973,x1974,x1975:length(x1963)=s(x1965) & leq(x1966, x1965)=true & (\/x1967,x1968,x1969,x1970,x1971:leq(x1966, x1965)=true & length(x1967)=x1965 ==> IF_6(x1968, x1969, x1970, x1971, x1967, x1966, true)_>=_IF_7(x1968, x1969, x1970, x1971, x1967, x1966, empty(fstsplit(x1966, x1967)))) ==> IF_6(x1972, x1973, x1974, x1975, x1963, s(x1966), true)_>=_IF_7(x1972, x1973, x1974, x1975, x1963, s(x1966), empty(fstsplit(s(x1966), x1963)))) ==> IF_6(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), true)_>=_IF_7(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), empty(fstsplit(s(x1957), cons(x1964, x1963))))) We simplified constraint (6) using rules (I), (II) which results in the following new constraint: (7) (length(x1963)=x1956 & leq(x1957, x1956)=true & (\/x1958,x1959,x1960,x1961,x1962:leq(x1957, x1956)=true & length(x1958)=x1956 ==> IF_6(x1959, x1960, x1961, x1962, x1958, x1957, true)_>=_IF_7(x1959, x1960, x1961, x1962, x1958, x1957, empty(fstsplit(x1957, x1958)))) & (\/x1965,x1966,x1967,x1968,x1969,x1970,x1971,x1972,x1973,x1974,x1975:length(x1963)=s(x1965) & leq(x1966, x1965)=true & (\/x1967,x1968,x1969,x1970,x1971:leq(x1966, x1965)=true & length(x1967)=x1965 ==> IF_6(x1968, x1969, x1970, x1971, x1967, x1966, true)_>=_IF_7(x1968, x1969, x1970, x1971, x1967, x1966, empty(fstsplit(x1966, x1967)))) ==> IF_6(x1972, x1973, x1974, x1975, x1963, s(x1966), true)_>=_IF_7(x1972, x1973, x1974, x1975, x1963, s(x1966), empty(fstsplit(s(x1966), x1963)))) ==> IF_6(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), true)_>=_IF_7(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), empty(fstsplit(s(x1957), cons(x1964, x1963))))) We simplified constraint (7) using rule (VI) where we applied the induction hypothesis (\/x1958,x1959,x1960,x1961,x1962:leq(x1957, x1956)=true & length(x1958)=x1956 ==> IF_6(x1959, x1960, x1961, x1962, x1958, x1957, true)_>=_IF_7(x1959, x1960, x1961, x1962, x1958, x1957, empty(fstsplit(x1957, x1958)))) with sigma = [x1958 / x1963, x1959 / x872, x1960 / x873, x1961 / x874, x1962 / x875] which results in the following new constraint: (8) (IF_6(x872, x873, x874, x875, x1963, x1957, true)_>=_IF_7(x872, x873, x874, x875, x1963, x1957, empty(fstsplit(x1957, x1963))) & (\/x1965,x1966,x1967,x1968,x1969,x1970,x1971,x1972,x1973,x1974,x1975:length(x1963)=s(x1965) & leq(x1966, x1965)=true & (\/x1967,x1968,x1969,x1970,x1971:leq(x1966, x1965)=true & length(x1967)=x1965 ==> IF_6(x1968, x1969, x1970, x1971, x1967, x1966, true)_>=_IF_7(x1968, x1969, x1970, x1971, x1967, x1966, empty(fstsplit(x1966, x1967)))) ==> IF_6(x1972, x1973, x1974, x1975, x1963, s(x1966), true)_>=_IF_7(x1972, x1973, x1974, x1975, x1963, s(x1966), empty(fstsplit(s(x1966), x1963)))) ==> IF_6(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), true)_>=_IF_7(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), empty(fstsplit(s(x1957), cons(x1964, x1963))))) We simplified constraint (8) using rule (IV) which results in the following new constraint: (9) (IF_6(x872, x873, x874, x875, x1963, x1957, true)_>=_IF_7(x872, x873, x874, x875, x1963, x1957, empty(fstsplit(x1957, x1963))) ==> IF_6(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), true)_>=_IF_7(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), empty(fstsplit(s(x1957), cons(x1964, x1963))))) For Pair IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) the following chains were created: *We consider the chain IF_6(x982, x983, x984, x985, x986, x987, true) -> IF_7(x982, x983, x984, x985, x986, x987, empty(fstsplit(x987, x986))), IF_7(x988, x989, x990, x991, x992, x993, false) -> RING(x988, x989, x990, x991, sndsplit(x993, x992), x993) which results in the following constraint: (1) (IF_7(x982, x983, x984, x985, x986, x987, empty(fstsplit(x987, x986)))=IF_7(x988, x989, x990, x991, x992, x993, false) ==> IF_7(x988, x989, x990, x991, x992, x993, false)_>=_RING(x988, x989, x990, x991, sndsplit(x993, x992), x993)) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (fstsplit(x987, x986)=x1976 & empty(x1976)=false ==> IF_7(x982, x983, x984, x985, x986, x987, false)_>=_RING(x982, x983, x984, x985, sndsplit(x987, x986), x987)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x1976)=false which results in the following new constraint: (3) (false=false & fstsplit(x987, x986)=cons(x1978, x1977) ==> IF_7(x982, x983, x984, x985, x986, x987, false)_>=_RING(x982, x983, x984, x985, sndsplit(x987, x986), x987)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (fstsplit(x987, x986)=cons(x1978, x1977) ==> IF_7(x982, x983, x984, x985, x986, x987, false)_>=_RING(x982, x983, x984, x985, sndsplit(x987, x986), x987)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on fstsplit(x987, x986)=cons(x1978, x1977) which results in the following new constraint: (5) (cons(x1982, fstsplit(x1983, x1981))=cons(x1978, x1977) & (\/x1984,x1985,x1986,x1987,x1988,x1989:fstsplit(x1983, x1981)=cons(x1984, x1985) ==> IF_7(x1986, x1987, x1988, x1989, x1981, x1983, false)_>=_RING(x1986, x1987, x1988, x1989, sndsplit(x1983, x1981), x1983)) ==> IF_7(x982, x983, x984, x985, cons(x1982, x1981), s(x1983), false)_>=_RING(x982, x983, x984, x985, sndsplit(s(x1983), cons(x1982, x1981)), s(x1983))) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (6) (IF_7(x982, x983, x984, x985, cons(x1982, x1981), s(x1983), false)_>=_RING(x982, x983, x984, x985, sndsplit(s(x1983), cons(x1982, x1981)), s(x1983))) For Pair IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) the following chains were created: *We consider the chain RING(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141) -> IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, empty(map_f(three, x1138))), IF_9(x1142, x1143, x1144, x1145, x1146, x1147, true) -> RING(x1142, x1143, x1144, tail(x1145), x1146, x1147) which results in the following constraint: (1) (IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, empty(map_f(three, x1138)))=IF_9(x1142, x1143, x1144, x1145, x1146, x1147, true) ==> IF_9(x1142, x1143, x1144, x1145, x1146, x1147, true)_>=_RING(x1142, x1143, x1144, tail(x1145), x1146, x1147)) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (three=x1991 & map_f(x1991, x1138)=x1990 & empty(x1990)=true ==> IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(x1138, x1139)), x1140, x1141)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x1990)=true which results in the following new constraint: (3) (true=true & three=x1991 & map_f(x1991, x1138)=nil ==> IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(x1138, x1139)), x1140, x1141)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (three=x1991 & map_f(x1991, x1138)=nil ==> IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(x1138, x1139)), x1140, x1141)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on map_f(x1991, x1138)=nil which results in the following new constraints: (5) (nil=nil & three=x1994 ==> IF_9(x1135, x1136, x1137, cons(nil, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(nil, x1139)), x1140, x1141)) (6) (app(f(x1997, x1996), map_f(x1997, x1995))=nil & three=x1997 & (\/x1998,x1999,x2000,x2001,x2002,x2003:map_f(x1997, x1995)=nil & three=x1997 ==> IF_9(x1998, x1999, x2000, cons(x1995, x2001), x2002, x2003, true)_>=_RING(x1998, x1999, x2000, tail(cons(x1995, x2001)), x2002, x2003)) ==> IF_9(x1135, x1136, x1137, cons(cons(x1996, x1995), x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(cons(x1996, x1995), x1139)), x1140, x1141)) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (7) (IF_9(x1135, x1136, x1137, cons(nil, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(nil, x1139)), x1140, x1141)) We simplified constraint (6) using rules (IV), (VII) which results in the following new constraint: (8) (f(x1997, x1996)=x2004 & map_f(x1997, x1995)=x2005 & app(x2004, x2005)=nil & three=x1997 ==> IF_9(x1135, x1136, x1137, cons(cons(x1996, x1995), x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(cons(x1996, x1995), x1139)), x1140, x1141)) We simplified constraint (8) using rule (V) (with possible (I) afterwards) using induction on app(x2004, x2005)=nil which results in the following new constraint: (9) (x2006=nil & f(x1997, x1996)=nil & map_f(x1997, x1995)=x2006 & three=x1997 ==> IF_9(x1135, x1136, x1137, cons(cons(x1996, x1995), x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(cons(x1996, x1995), x1139)), x1140, x1141)) We solved constraint (9) using rules (I), (II). For Pair IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) the following chains were created: *We consider the chain RING(x1184, x1185, x1186, x1187, x1188, x1189) -> IF_6(x1184, x1185, x1186, x1187, x1188, x1189, leq(x1189, length(x1188))), IF_6(x1190, x1191, x1192, x1193, x1194, x1195, false) -> IF_8(x1190, x1191, x1192, x1193, x1194, x1195, empty(fstsplit(x1195, app(map_f(three, head(x1193)), x1194)))) which results in the following constraint: (1) (IF_6(x1184, x1185, x1186, x1187, x1188, x1189, leq(x1189, length(x1188)))=IF_6(x1190, x1191, x1192, x1193, x1194, x1195, false) ==> IF_6(x1190, x1191, x1192, x1193, x1194, x1195, false)_>=_IF_8(x1190, x1191, x1192, x1193, x1194, x1195, empty(fstsplit(x1195, app(map_f(three, head(x1193)), x1194))))) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (length(x1188)=x2010 & leq(x1189, x2010)=false ==> IF_6(x1184, x1185, x1186, x1187, x1188, x1189, false)_>=_IF_8(x1184, x1185, x1186, x1187, x1188, x1189, empty(fstsplit(x1189, app(map_f(three, head(x1187)), x1188))))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on leq(x1189, x2010)=false which results in the following new constraints: (3) (false=false & length(x1188)=0 ==> IF_6(x1184, x1185, x1186, x1187, x1188, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, x1188, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), x1188))))) (4) (leq(x2014, x2013)=false & length(x1188)=s(x2013) & (\/x2015,x2016,x2017,x2018,x2019:leq(x2014, x2013)=false & length(x2015)=x2013 ==> IF_6(x2016, x2017, x2018, x2019, x2015, x2014, false)_>=_IF_8(x2016, x2017, x2018, x2019, x2015, x2014, empty(fstsplit(x2014, app(map_f(three, head(x2019)), x2015))))) ==> IF_6(x1184, x1185, x1186, x1187, x1188, s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, x1188, s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), x1188))))) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (5) (length(x1188)=0 ==> IF_6(x1184, x1185, x1186, x1187, x1188, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, x1188, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), x1188))))) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on length(x1188)=s(x2013) which results in the following new constraint: (6) (s(length(x2022))=s(x2013) & leq(x2014, x2013)=false & (\/x2015,x2016,x2017,x2018,x2019:leq(x2014, x2013)=false & length(x2015)=x2013 ==> IF_6(x2016, x2017, x2018, x2019, x2015, x2014, false)_>=_IF_8(x2016, x2017, x2018, x2019, x2015, x2014, empty(fstsplit(x2014, app(map_f(three, head(x2019)), x2015))))) & (\/x2024,x2025,x2026,x2027,x2028,x2029,x2030,x2031,x2032,x2033,x2034:length(x2022)=s(x2024) & leq(x2025, x2024)=false & (\/x2026,x2027,x2028,x2029,x2030:leq(x2025, x2024)=false & length(x2026)=x2024 ==> IF_6(x2027, x2028, x2029, x2030, x2026, x2025, false)_>=_IF_8(x2027, x2028, x2029, x2030, x2026, x2025, empty(fstsplit(x2025, app(map_f(three, head(x2030)), x2026))))) ==> IF_6(x2031, x2032, x2033, x2034, x2022, s(x2025), false)_>=_IF_8(x2031, x2032, x2033, x2034, x2022, s(x2025), empty(fstsplit(s(x2025), app(map_f(three, head(x2034)), x2022))))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on length(x1188)=0 which results in the following new constraint: (7) (0=0 ==> IF_6(x1184, x1185, x1186, x1187, nil, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, nil, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), nil))))) We simplified constraint (7) using rules (I), (II) which results in the following new constraint: (8) (IF_6(x1184, x1185, x1186, x1187, nil, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, nil, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), nil))))) We simplified constraint (6) using rules (I), (II) which results in the following new constraint: (9) (length(x2022)=x2013 & leq(x2014, x2013)=false & (\/x2015,x2016,x2017,x2018,x2019:leq(x2014, x2013)=false & length(x2015)=x2013 ==> IF_6(x2016, x2017, x2018, x2019, x2015, x2014, false)_>=_IF_8(x2016, x2017, x2018, x2019, x2015, x2014, empty(fstsplit(x2014, app(map_f(three, head(x2019)), x2015))))) & (\/x2024,x2025,x2026,x2027,x2028,x2029,x2030,x2031,x2032,x2033,x2034:length(x2022)=s(x2024) & leq(x2025, x2024)=false & (\/x2026,x2027,x2028,x2029,x2030:leq(x2025, x2024)=false & length(x2026)=x2024 ==> IF_6(x2027, x2028, x2029, x2030, x2026, x2025, false)_>=_IF_8(x2027, x2028, x2029, x2030, x2026, x2025, empty(fstsplit(x2025, app(map_f(three, head(x2030)), x2026))))) ==> IF_6(x2031, x2032, x2033, x2034, x2022, s(x2025), false)_>=_IF_8(x2031, x2032, x2033, x2034, x2022, s(x2025), empty(fstsplit(s(x2025), app(map_f(three, head(x2034)), x2022))))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) We simplified constraint (9) using rule (VI) where we applied the induction hypothesis (\/x2015,x2016,x2017,x2018,x2019:leq(x2014, x2013)=false & length(x2015)=x2013 ==> IF_6(x2016, x2017, x2018, x2019, x2015, x2014, false)_>=_IF_8(x2016, x2017, x2018, x2019, x2015, x2014, empty(fstsplit(x2014, app(map_f(three, head(x2019)), x2015))))) with sigma = [x2015 / x2022, x2016 / x1184, x2017 / x1185, x2018 / x1186, x2019 / x1187] which results in the following new constraint: (10) (IF_6(x1184, x1185, x1186, x1187, x2022, x2014, false)_>=_IF_8(x1184, x1185, x1186, x1187, x2022, x2014, empty(fstsplit(x2014, app(map_f(three, head(x1187)), x2022)))) & (\/x2024,x2025,x2026,x2027,x2028,x2029,x2030,x2031,x2032,x2033,x2034:length(x2022)=s(x2024) & leq(x2025, x2024)=false & (\/x2026,x2027,x2028,x2029,x2030:leq(x2025, x2024)=false & length(x2026)=x2024 ==> IF_6(x2027, x2028, x2029, x2030, x2026, x2025, false)_>=_IF_8(x2027, x2028, x2029, x2030, x2026, x2025, empty(fstsplit(x2025, app(map_f(three, head(x2030)), x2026))))) ==> IF_6(x2031, x2032, x2033, x2034, x2022, s(x2025), false)_>=_IF_8(x2031, x2032, x2033, x2034, x2022, s(x2025), empty(fstsplit(s(x2025), app(map_f(three, head(x2034)), x2022))))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) We simplified constraint (10) using rule (IV) which results in the following new constraint: (11) (IF_6(x1184, x1185, x1186, x1187, x2022, x2014, false)_>=_IF_8(x1184, x1185, x1186, x1187, x2022, x2014, empty(fstsplit(x2014, app(map_f(three, head(x1187)), x2022)))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) For Pair IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) the following chains were created: *We consider the chain IF_6(x1312, x1313, x1314, x1315, x1316, x1317, false) -> IF_8(x1312, x1313, x1314, x1315, x1316, x1317, empty(fstsplit(x1317, app(map_f(three, head(x1315)), x1316)))), IF_8(x1318, x1319, x1320, x1321, x1322, x1323, false) -> RING(x1318, x1319, x1320, tail(x1321), sndsplit(x1323, app(map_f(three, head(x1321)), x1322)), x1323) which results in the following constraint: (1) (IF_8(x1312, x1313, x1314, x1315, x1316, x1317, empty(fstsplit(x1317, app(map_f(three, head(x1315)), x1316))))=IF_8(x1318, x1319, x1320, x1321, x1322, x1323, false) ==> IF_8(x1318, x1319, x1320, x1321, x1322, x1323, false)_>=_RING(x1318, x1319, x1320, tail(x1321), sndsplit(x1323, app(map_f(three, head(x1321)), x1322)), x1323)) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (fstsplit(x1317, x2036)=x2035 & empty(x2035)=false ==> IF_8(x1312, x1313, x1314, x1315, x1316, x1317, false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(x1317, app(map_f(three, head(x1315)), x1316)), x1317)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x2035)=false which results in the following new constraint: (3) (false=false & fstsplit(x1317, x2036)=cons(x2041, x2040) ==> IF_8(x1312, x1313, x1314, x1315, x1316, x1317, false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(x1317, app(map_f(three, head(x1315)), x1316)), x1317)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (fstsplit(x1317, x2036)=cons(x2041, x2040) ==> IF_8(x1312, x1313, x1314, x1315, x1316, x1317, false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(x1317, app(map_f(three, head(x1315)), x1316)), x1317)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on fstsplit(x1317, x2036)=cons(x2041, x2040) which results in the following new constraint: (5) (cons(x2045, fstsplit(x2046, x2044))=cons(x2041, x2040) & (\/x2047,x2048,x2049,x2050,x2051,x2052,x2053:fstsplit(x2046, x2044)=cons(x2047, x2048) ==> IF_8(x2049, x2050, x2051, x2052, x2053, x2046, false)_>=_RING(x2049, x2050, x2051, tail(x2052), sndsplit(x2046, app(map_f(three, head(x2052)), x2053)), x2046)) ==> IF_8(x1312, x1313, x1314, x1315, x1316, s(x2046), false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(s(x2046), app(map_f(three, head(x1315)), x1316)), s(x2046))) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (6) (IF_8(x1312, x1313, x1314, x1315, x1316, s(x2046), false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(s(x2046), app(map_f(three, head(x1315)), x1316)), s(x2046))) For Pair IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) the following chains were created: *We consider the chain RING(x1368, x1369, x1370, x1371, x1372, x1373) -> IF_2(x1368, x1369, x1370, x1371, x1372, x1373, leq(x1373, length(x1370))), IF_2(x1374, x1375, x1376, x1377, x1378, x1379, false) -> IF_4(x1374, x1375, x1376, x1377, x1378, x1379, empty(fstsplit(x1379, app(map_f(two, head(x1375)), x1376)))) which results in the following constraint: (1) (IF_2(x1368, x1369, x1370, x1371, x1372, x1373, leq(x1373, length(x1370)))=IF_2(x1374, x1375, x1376, x1377, x1378, x1379, false) ==> IF_2(x1374, x1375, x1376, x1377, x1378, x1379, false)_>=_IF_4(x1374, x1375, x1376, x1377, x1378, x1379, empty(fstsplit(x1379, app(map_f(two, head(x1375)), x1376))))) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (length(x1370)=x2054 & leq(x1373, x2054)=false ==> IF_2(x1368, x1369, x1370, x1371, x1372, x1373, false)_>=_IF_4(x1368, x1369, x1370, x1371, x1372, x1373, empty(fstsplit(x1373, app(map_f(two, head(x1369)), x1370))))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on leq(x1373, x2054)=false which results in the following new constraints: (3) (false=false & length(x1370)=0 ==> IF_2(x1368, x1369, x1370, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, x1370, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), x1370))))) (4) (leq(x2058, x2057)=false & length(x1370)=s(x2057) & (\/x2059,x2060,x2061,x2062,x2063:leq(x2058, x2057)=false & length(x2059)=x2057 ==> IF_2(x2060, x2061, x2059, x2062, x2063, x2058, false)_>=_IF_4(x2060, x2061, x2059, x2062, x2063, x2058, empty(fstsplit(x2058, app(map_f(two, head(x2061)), x2059))))) ==> IF_2(x1368, x1369, x1370, x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, x1370, x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), x1370))))) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (5) (length(x1370)=0 ==> IF_2(x1368, x1369, x1370, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, x1370, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), x1370))))) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on length(x1370)=s(x2057) which results in the following new constraint: (6) (s(length(x2066))=s(x2057) & leq(x2058, x2057)=false & (\/x2059,x2060,x2061,x2062,x2063:leq(x2058, x2057)=false & length(x2059)=x2057 ==> IF_2(x2060, x2061, x2059, x2062, x2063, x2058, false)_>=_IF_4(x2060, x2061, x2059, x2062, x2063, x2058, empty(fstsplit(x2058, app(map_f(two, head(x2061)), x2059))))) & (\/x2068,x2069,x2070,x2071,x2072,x2073,x2074,x2075,x2076,x2077,x2078:length(x2066)=s(x2068) & leq(x2069, x2068)=false & (\/x2070,x2071,x2072,x2073,x2074:leq(x2069, x2068)=false & length(x2070)=x2068 ==> IF_2(x2071, x2072, x2070, x2073, x2074, x2069, false)_>=_IF_4(x2071, x2072, x2070, x2073, x2074, x2069, empty(fstsplit(x2069, app(map_f(two, head(x2072)), x2070))))) ==> IF_2(x2075, x2076, x2066, x2077, x2078, s(x2069), false)_>=_IF_4(x2075, x2076, x2066, x2077, x2078, s(x2069), empty(fstsplit(s(x2069), app(map_f(two, head(x2076)), x2066))))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on length(x1370)=0 which results in the following new constraint: (7) (0=0 ==> IF_2(x1368, x1369, nil, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, nil, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), nil))))) We simplified constraint (7) using rules (I), (II) which results in the following new constraint: (8) (IF_2(x1368, x1369, nil, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, nil, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), nil))))) We simplified constraint (6) using rules (I), (II) which results in the following new constraint: (9) (length(x2066)=x2057 & leq(x2058, x2057)=false & (\/x2059,x2060,x2061,x2062,x2063:leq(x2058, x2057)=false & length(x2059)=x2057 ==> IF_2(x2060, x2061, x2059, x2062, x2063, x2058, false)_>=_IF_4(x2060, x2061, x2059, x2062, x2063, x2058, empty(fstsplit(x2058, app(map_f(two, head(x2061)), x2059))))) & (\/x2068,x2069,x2070,x2071,x2072,x2073,x2074,x2075,x2076,x2077,x2078:length(x2066)=s(x2068) & leq(x2069, x2068)=false & (\/x2070,x2071,x2072,x2073,x2074:leq(x2069, x2068)=false & length(x2070)=x2068 ==> IF_2(x2071, x2072, x2070, x2073, x2074, x2069, false)_>=_IF_4(x2071, x2072, x2070, x2073, x2074, x2069, empty(fstsplit(x2069, app(map_f(two, head(x2072)), x2070))))) ==> IF_2(x2075, x2076, x2066, x2077, x2078, s(x2069), false)_>=_IF_4(x2075, x2076, x2066, x2077, x2078, s(x2069), empty(fstsplit(s(x2069), app(map_f(two, head(x2076)), x2066))))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) We simplified constraint (9) using rule (VI) where we applied the induction hypothesis (\/x2059,x2060,x2061,x2062,x2063:leq(x2058, x2057)=false & length(x2059)=x2057 ==> IF_2(x2060, x2061, x2059, x2062, x2063, x2058, false)_>=_IF_4(x2060, x2061, x2059, x2062, x2063, x2058, empty(fstsplit(x2058, app(map_f(two, head(x2061)), x2059))))) with sigma = [x2059 / x2066, x2060 / x1368, x2061 / x1369, x2062 / x1371, x2063 / x1372] which results in the following new constraint: (10) (IF_2(x1368, x1369, x2066, x1371, x1372, x2058, false)_>=_IF_4(x1368, x1369, x2066, x1371, x1372, x2058, empty(fstsplit(x2058, app(map_f(two, head(x1369)), x2066)))) & (\/x2068,x2069,x2070,x2071,x2072,x2073,x2074,x2075,x2076,x2077,x2078:length(x2066)=s(x2068) & leq(x2069, x2068)=false & (\/x2070,x2071,x2072,x2073,x2074:leq(x2069, x2068)=false & length(x2070)=x2068 ==> IF_2(x2071, x2072, x2070, x2073, x2074, x2069, false)_>=_IF_4(x2071, x2072, x2070, x2073, x2074, x2069, empty(fstsplit(x2069, app(map_f(two, head(x2072)), x2070))))) ==> IF_2(x2075, x2076, x2066, x2077, x2078, s(x2069), false)_>=_IF_4(x2075, x2076, x2066, x2077, x2078, s(x2069), empty(fstsplit(s(x2069), app(map_f(two, head(x2076)), x2066))))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) We simplified constraint (10) using rule (IV) which results in the following new constraint: (11) (IF_2(x1368, x1369, x2066, x1371, x1372, x2058, false)_>=_IF_4(x1368, x1369, x2066, x1371, x1372, x2058, empty(fstsplit(x2058, app(map_f(two, head(x1369)), x2066)))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) For Pair IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) the following chains were created: *We consider the chain IF_2(x1532, x1533, x1534, x1535, x1536, x1537, false) -> IF_4(x1532, x1533, x1534, x1535, x1536, x1537, empty(fstsplit(x1537, app(map_f(two, head(x1533)), x1534)))), IF_4(x1538, x1539, x1540, x1541, x1542, x1543, false) -> RING(x1538, tail(x1539), sndsplit(x1543, app(map_f(two, head(x1539)), x1540)), cons(fstsplit(x1543, app(map_f(two, head(x1539)), x1540)), x1541), x1542, x1543) which results in the following constraint: (1) (IF_4(x1532, x1533, x1534, x1535, x1536, x1537, empty(fstsplit(x1537, app(map_f(two, head(x1533)), x1534))))=IF_4(x1538, x1539, x1540, x1541, x1542, x1543, false) ==> IF_4(x1538, x1539, x1540, x1541, x1542, x1543, false)_>=_RING(x1538, tail(x1539), sndsplit(x1543, app(map_f(two, head(x1539)), x1540)), cons(fstsplit(x1543, app(map_f(two, head(x1539)), x1540)), x1541), x1542, x1543)) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (fstsplit(x1537, x2080)=x2079 & empty(x2079)=false ==> IF_4(x1532, x1533, x1534, x1535, x1536, x1537, false)_>=_RING(x1532, tail(x1533), sndsplit(x1537, app(map_f(two, head(x1533)), x1534)), cons(fstsplit(x1537, app(map_f(two, head(x1533)), x1534)), x1535), x1536, x1537)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x2079)=false which results in the following new constraint: (3) (false=false & fstsplit(x1537, x2080)=cons(x2085, x2084) ==> IF_4(x1532, x1533, x1534, x1535, x1536, x1537, false)_>=_RING(x1532, tail(x1533), sndsplit(x1537, app(map_f(two, head(x1533)), x1534)), cons(fstsplit(x1537, app(map_f(two, head(x1533)), x1534)), x1535), x1536, x1537)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (fstsplit(x1537, x2080)=cons(x2085, x2084) ==> IF_4(x1532, x1533, x1534, x1535, x1536, x1537, false)_>=_RING(x1532, tail(x1533), sndsplit(x1537, app(map_f(two, head(x1533)), x1534)), cons(fstsplit(x1537, app(map_f(two, head(x1533)), x1534)), x1535), x1536, x1537)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on fstsplit(x1537, x2080)=cons(x2085, x2084) which results in the following new constraint: (5) (cons(x2089, fstsplit(x2090, x2088))=cons(x2085, x2084) & (\/x2091,x2092,x2093,x2094,x2095,x2096,x2097:fstsplit(x2090, x2088)=cons(x2091, x2092) ==> IF_4(x2093, x2094, x2095, x2096, x2097, x2090, false)_>=_RING(x2093, tail(x2094), sndsplit(x2090, app(map_f(two, head(x2094)), x2095)), cons(fstsplit(x2090, app(map_f(two, head(x2094)), x2095)), x2096), x2097, x2090)) ==> IF_4(x1532, x1533, x1534, x1535, x1536, s(x2090), false)_>=_RING(x1532, tail(x1533), sndsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), cons(fstsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), x1535), x1536, s(x2090))) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (6) (IF_4(x1532, x1533, x1534, x1535, x1536, s(x2090), false)_>=_RING(x1532, tail(x1533), sndsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), cons(fstsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), x1535), x1536, s(x2090))) For Pair RING(y0, cons(x0, x1), y2, y3, y4, y5) -> IF_5(y0, cons(x0, x1), y2, y3, y4, y5, empty(map_f(two, x0))) the following chains were created: *We consider the chain IF_1(x1570, x1571, x1572, x1573, x1574, x1575, false) -> RING(sndsplit(x1575, x1570), cons(fstsplit(x1575, x1570), x1571), x1572, x1573, x1574, x1575), RING(x1576, cons(x1577, x1578), x1579, x1580, x1581, x1582) -> IF_5(x1576, cons(x1577, x1578), x1579, x1580, x1581, x1582, empty(map_f(two, x1577))) which results in the following constraint: (1) (RING(sndsplit(x1575, x1570), cons(fstsplit(x1575, x1570), x1571), x1572, x1573, x1574, x1575)=RING(x1576, cons(x1577, x1578), x1579, x1580, x1581, x1582) ==> RING(x1576, cons(x1577, x1578), x1579, x1580, x1581, x1582)_>=_IF_5(x1576, cons(x1577, x1578), x1579, x1580, x1581, x1582, empty(map_f(two, x1577)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x1576, cons(x1577, x1571), x1572, x1573, x1574, x1575)_>=_IF_5(x1576, cons(x1577, x1571), x1572, x1573, x1574, x1575, empty(map_f(two, x1577)))) *We consider the chain IF_3(x1595, x1596, x1597, x1598, x1599, x1600, false) -> RING(x1595, x1596, sndsplit(x1600, x1597), cons(fstsplit(x1600, x1597), x1598), x1599, x1600), RING(x1601, cons(x1602, x1603), x1604, x1605, x1606, x1607) -> IF_5(x1601, cons(x1602, x1603), x1604, x1605, x1606, x1607, empty(map_f(two, x1602))) which results in the following constraint: (1) (RING(x1595, x1596, sndsplit(x1600, x1597), cons(fstsplit(x1600, x1597), x1598), x1599, x1600)=RING(x1601, cons(x1602, x1603), x1604, x1605, x1606, x1607) ==> RING(x1601, cons(x1602, x1603), x1604, x1605, x1606, x1607)_>=_IF_5(x1601, cons(x1602, x1603), x1604, x1605, x1606, x1607, empty(map_f(two, x1602)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x1595, cons(x1602, x1603), x1604, x1605, x1599, x1600)_>=_IF_5(x1595, cons(x1602, x1603), x1604, x1605, x1599, x1600, empty(map_f(two, x1602)))) *We consider the chain IF_5(x1608, x1609, x1610, x1611, x1612, x1613, true) -> RING(x1608, tail(x1609), x1610, x1611, x1612, x1613), RING(x1614, cons(x1615, x1616), x1617, x1618, x1619, x1620) -> IF_5(x1614, cons(x1615, x1616), x1617, x1618, x1619, x1620, empty(map_f(two, x1615))) which results in the following constraint: (1) (RING(x1608, tail(x1609), x1610, x1611, x1612, x1613)=RING(x1614, cons(x1615, x1616), x1617, x1618, x1619, x1620) ==> RING(x1614, cons(x1615, x1616), x1617, x1618, x1619, x1620)_>=_IF_5(x1614, cons(x1615, x1616), x1617, x1618, x1619, x1620, empty(map_f(two, x1615)))) We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint: (2) (tail(x1609)=cons(x1615, x1616) ==> RING(x1608, cons(x1615, x1616), x1610, x1611, x1612, x1613)_>=_IF_5(x1608, cons(x1615, x1616), x1610, x1611, x1612, x1613, empty(map_f(two, x1615)))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on tail(x1609)=cons(x1615, x1616) which results in the following new constraint: (3) (x2098=cons(x1615, x1616) ==> RING(x1608, cons(x1615, x1616), x1610, x1611, x1612, x1613)_>=_IF_5(x1608, cons(x1615, x1616), x1610, x1611, x1612, x1613, empty(map_f(two, x1615)))) We simplified constraint (3) using rule (III) which results in the following new constraint: (4) (RING(x1608, cons(x1615, x1616), x1610, x1611, x1612, x1613)_>=_IF_5(x1608, cons(x1615, x1616), x1610, x1611, x1612, x1613, empty(map_f(two, x1615)))) *We consider the chain IF_7(x1633, x1634, x1635, x1636, x1637, x1638, false) -> RING(x1633, x1634, x1635, x1636, sndsplit(x1638, x1637), x1638), RING(x1639, cons(x1640, x1641), x1642, x1643, x1644, x1645) -> IF_5(x1639, cons(x1640, x1641), x1642, x1643, x1644, x1645, empty(map_f(two, x1640))) which results in the following constraint: (1) (RING(x1633, x1634, x1635, x1636, sndsplit(x1638, x1637), x1638)=RING(x1639, cons(x1640, x1641), x1642, x1643, x1644, x1645) ==> RING(x1639, cons(x1640, x1641), x1642, x1643, x1644, x1645)_>=_IF_5(x1639, cons(x1640, x1641), x1642, x1643, x1644, x1645, empty(map_f(two, x1640)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x1633, cons(x1640, x1641), x1635, x1636, x1644, x1638)_>=_IF_5(x1633, cons(x1640, x1641), x1635, x1636, x1644, x1638, empty(map_f(two, x1640)))) *We consider the chain IF_9(x1646, x1647, x1648, x1649, x1650, x1651, true) -> RING(x1646, x1647, x1648, tail(x1649), x1650, x1651), RING(x1652, cons(x1653, x1654), x1655, x1656, x1657, x1658) -> IF_5(x1652, cons(x1653, x1654), x1655, x1656, x1657, x1658, empty(map_f(two, x1653))) which results in the following constraint: (1) (RING(x1646, x1647, x1648, tail(x1649), x1650, x1651)=RING(x1652, cons(x1653, x1654), x1655, x1656, x1657, x1658) ==> RING(x1652, cons(x1653, x1654), x1655, x1656, x1657, x1658)_>=_IF_5(x1652, cons(x1653, x1654), x1655, x1656, x1657, x1658, empty(map_f(two, x1653)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x1646, cons(x1653, x1654), x1648, x1656, x1650, x1651)_>=_IF_5(x1646, cons(x1653, x1654), x1648, x1656, x1650, x1651, empty(map_f(two, x1653)))) *We consider the chain IF_8(x1665, x1666, x1667, x1668, x1669, x1670, false) -> RING(x1665, x1666, x1667, tail(x1668), sndsplit(x1670, app(map_f(three, head(x1668)), x1669)), x1670), RING(x1671, cons(x1672, x1673), x1674, x1675, x1676, x1677) -> IF_5(x1671, cons(x1672, x1673), x1674, x1675, x1676, x1677, empty(map_f(two, x1672))) which results in the following constraint: (1) (RING(x1665, x1666, x1667, tail(x1668), sndsplit(x1670, app(map_f(three, head(x1668)), x1669)), x1670)=RING(x1671, cons(x1672, x1673), x1674, x1675, x1676, x1677) ==> RING(x1671, cons(x1672, x1673), x1674, x1675, x1676, x1677)_>=_IF_5(x1671, cons(x1672, x1673), x1674, x1675, x1676, x1677, empty(map_f(two, x1672)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x1665, cons(x1672, x1673), x1667, x1675, x1676, x1670)_>=_IF_5(x1665, cons(x1672, x1673), x1667, x1675, x1676, x1670, empty(map_f(two, x1672)))) *We consider the chain IF_4(x1684, x1685, x1686, x1687, x1688, x1689, false) -> RING(x1684, tail(x1685), sndsplit(x1689, app(map_f(two, head(x1685)), x1686)), cons(fstsplit(x1689, app(map_f(two, head(x1685)), x1686)), x1687), x1688, x1689), RING(x1690, cons(x1691, x1692), x1693, x1694, x1695, x1696) -> IF_5(x1690, cons(x1691, x1692), x1693, x1694, x1695, x1696, empty(map_f(two, x1691))) which results in the following constraint: (1) (RING(x1684, tail(x1685), sndsplit(x1689, app(map_f(two, head(x1685)), x1686)), cons(fstsplit(x1689, app(map_f(two, head(x1685)), x1686)), x1687), x1688, x1689)=RING(x1690, cons(x1691, x1692), x1693, x1694, x1695, x1696) ==> RING(x1690, cons(x1691, x1692), x1693, x1694, x1695, x1696)_>=_IF_5(x1690, cons(x1691, x1692), x1693, x1694, x1695, x1696, empty(map_f(two, x1691)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (tail(x1685)=cons(x1691, x1692) & app(x2105, x1686)=x2104 & sndsplit(x1689, x2104)=x1693 & cons(fstsplit(x1689, app(map_f(two, head(x1685)), x1686)), x1687)=x1694 ==> RING(x1684, cons(x1691, x1692), x1693, x1694, x1688, x1689)_>=_IF_5(x1684, cons(x1691, x1692), x1693, x1694, x1688, x1689, empty(map_f(two, x1691)))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on tail(x1685)=cons(x1691, x1692) which results in the following new constraint: (3) (x2108=cons(x1691, x1692) & app(x2105, x1686)=x2104 & sndsplit(x1689, x2104)=x1693 & cons(fstsplit(x1689, app(map_f(two, head(cons(x2109, x2108))), x1686)), x1687)=x1694 ==> RING(x1684, cons(x1691, x1692), x1693, x1694, x1688, x1689)_>=_IF_5(x1684, cons(x1691, x1692), x1693, x1694, x1688, x1689, empty(map_f(two, x1691)))) We simplified constraint (3) using rules (III), (IV) which results in the following new constraint: (4) (RING(x1684, cons(x1691, x1692), x1693, x1694, x1688, x1689)_>=_IF_5(x1684, cons(x1691, x1692), x1693, x1694, x1688, x1689, empty(map_f(two, x1691)))) For Pair RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) the following chains were created: *We consider the chain IF_1(x1717, x1718, x1719, x1720, x1721, x1722, false) -> RING(sndsplit(x1722, x1717), cons(fstsplit(x1722, x1717), x1718), x1719, x1720, x1721, x1722), RING(x1723, x1724, x1725, cons(x1726, x1727), x1728, x1729) -> IF_9(x1723, x1724, x1725, cons(x1726, x1727), x1728, x1729, empty(map_f(three, x1726))) which results in the following constraint: (1) (RING(sndsplit(x1722, x1717), cons(fstsplit(x1722, x1717), x1718), x1719, x1720, x1721, x1722)=RING(x1723, x1724, x1725, cons(x1726, x1727), x1728, x1729) ==> RING(x1723, x1724, x1725, cons(x1726, x1727), x1728, x1729)_>=_IF_9(x1723, x1724, x1725, cons(x1726, x1727), x1728, x1729, empty(map_f(three, x1726)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x1723, x1724, x1719, cons(x1726, x1727), x1721, x1722)_>=_IF_9(x1723, x1724, x1719, cons(x1726, x1727), x1721, x1722, empty(map_f(three, x1726)))) *We consider the chain IF_3(x1742, x1743, x1744, x1745, x1746, x1747, false) -> RING(x1742, x1743, sndsplit(x1747, x1744), cons(fstsplit(x1747, x1744), x1745), x1746, x1747), RING(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754) -> IF_9(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754, empty(map_f(three, x1751))) which results in the following constraint: (1) (RING(x1742, x1743, sndsplit(x1747, x1744), cons(fstsplit(x1747, x1744), x1745), x1746, x1747)=RING(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754) ==> RING(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754)_>=_IF_9(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754, empty(map_f(three, x1751)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x1742, x1743, x1750, cons(x1751, x1745), x1746, x1747)_>=_IF_9(x1742, x1743, x1750, cons(x1751, x1745), x1746, x1747, empty(map_f(three, x1751)))) *We consider the chain IF_5(x1755, x1756, x1757, x1758, x1759, x1760, true) -> RING(x1755, tail(x1756), x1757, x1758, x1759, x1760), RING(x1761, x1762, x1763, cons(x1764, x1765), x1766, x1767) -> IF_9(x1761, x1762, x1763, cons(x1764, x1765), x1766, x1767, empty(map_f(three, x1764))) which results in the following constraint: (1) (RING(x1755, tail(x1756), x1757, x1758, x1759, x1760)=RING(x1761, x1762, x1763, cons(x1764, x1765), x1766, x1767) ==> RING(x1761, x1762, x1763, cons(x1764, x1765), x1766, x1767)_>=_IF_9(x1761, x1762, x1763, cons(x1764, x1765), x1766, x1767, empty(map_f(three, x1764)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x1755, x1762, x1757, cons(x1764, x1765), x1759, x1760)_>=_IF_9(x1755, x1762, x1757, cons(x1764, x1765), x1759, x1760, empty(map_f(three, x1764)))) *We consider the chain IF_7(x1780, x1781, x1782, x1783, x1784, x1785, false) -> RING(x1780, x1781, x1782, x1783, sndsplit(x1785, x1784), x1785), RING(x1786, x1787, x1788, cons(x1789, x1790), x1791, x1792) -> IF_9(x1786, x1787, x1788, cons(x1789, x1790), x1791, x1792, empty(map_f(three, x1789))) which results in the following constraint: (1) (RING(x1780, x1781, x1782, x1783, sndsplit(x1785, x1784), x1785)=RING(x1786, x1787, x1788, cons(x1789, x1790), x1791, x1792) ==> RING(x1786, x1787, x1788, cons(x1789, x1790), x1791, x1792)_>=_IF_9(x1786, x1787, x1788, cons(x1789, x1790), x1791, x1792, empty(map_f(three, x1789)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x1780, x1781, x1782, cons(x1789, x1790), x1791, x1785)_>=_IF_9(x1780, x1781, x1782, cons(x1789, x1790), x1791, x1785, empty(map_f(three, x1789)))) *We consider the chain IF_9(x1793, x1794, x1795, x1796, x1797, x1798, true) -> RING(x1793, x1794, x1795, tail(x1796), x1797, x1798), RING(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805) -> IF_9(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805, empty(map_f(three, x1802))) which results in the following constraint: (1) (RING(x1793, x1794, x1795, tail(x1796), x1797, x1798)=RING(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805) ==> RING(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805)_>=_IF_9(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805, empty(map_f(three, x1802)))) We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint: (2) (tail(x1796)=cons(x1802, x1803) ==> RING(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798)_>=_IF_9(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798, empty(map_f(three, x1802)))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on tail(x1796)=cons(x1802, x1803) which results in the following new constraint: (3) (x2110=cons(x1802, x1803) ==> RING(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798)_>=_IF_9(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798, empty(map_f(three, x1802)))) We simplified constraint (3) using rule (III) which results in the following new constraint: (4) (RING(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798)_>=_IF_9(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798, empty(map_f(three, x1802)))) *We consider the chain IF_8(x1812, x1813, x1814, x1815, x1816, x1817, false) -> RING(x1812, x1813, x1814, tail(x1815), sndsplit(x1817, app(map_f(three, head(x1815)), x1816)), x1817), RING(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824) -> IF_9(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824, empty(map_f(three, x1821))) which results in the following constraint: (1) (RING(x1812, x1813, x1814, tail(x1815), sndsplit(x1817, app(map_f(three, head(x1815)), x1816)), x1817)=RING(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824) ==> RING(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824)_>=_IF_9(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824, empty(map_f(three, x1821)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (tail(x1815)=cons(x1821, x1822) & app(x2113, x1816)=x2112 & sndsplit(x1817, x2112)=x1823 ==> RING(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817)_>=_IF_9(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817, empty(map_f(three, x1821)))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on tail(x1815)=cons(x1821, x1822) which results in the following new constraint: (3) (x2116=cons(x1821, x1822) & app(x2113, x1816)=x2112 & sndsplit(x1817, x2112)=x1823 ==> RING(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817)_>=_IF_9(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817, empty(map_f(three, x1821)))) We simplified constraint (3) using rules (III), (IV) which results in the following new constraint: (4) (RING(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817)_>=_IF_9(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817, empty(map_f(three, x1821)))) *We consider the chain IF_4(x1831, x1832, x1833, x1834, x1835, x1836, false) -> RING(x1831, tail(x1832), sndsplit(x1836, app(map_f(two, head(x1832)), x1833)), cons(fstsplit(x1836, app(map_f(two, head(x1832)), x1833)), x1834), x1835, x1836), RING(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843) -> IF_9(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843, empty(map_f(three, x1840))) which results in the following constraint: (1) (RING(x1831, tail(x1832), sndsplit(x1836, app(map_f(two, head(x1832)), x1833)), cons(fstsplit(x1836, app(map_f(two, head(x1832)), x1833)), x1834), x1835, x1836)=RING(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843) ==> RING(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843)_>=_IF_9(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843, empty(map_f(three, x1840)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x1831, x1838, x1839, cons(x1840, x1834), x1835, x1836)_>=_IF_9(x1831, x1838, x1839, cons(x1840, x1834), x1835, x1836, empty(map_f(three, x1840)))) To summarize, we get the following constraints P__>=_ for the following pairs. *RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) *(RING(x12, x13, x8, x9, x10, x11)_>=_IF_1(x12, x13, x8, x9, x10, x11, empty(fstsplit(x11, x12)))) *(RING(x30, x31, x38, x39, x34, x35)_>=_IF_1(x30, x31, x38, x39, x34, x35, empty(fstsplit(x35, x30)))) *(RING(x42, x49, x44, x45, x46, x47)_>=_IF_1(x42, x49, x44, x45, x46, x47, empty(fstsplit(x47, x42)))) *(RING(x66, x67, x68, x69, x76, x71)_>=_IF_1(x66, x67, x68, x69, x76, x71, empty(fstsplit(x71, x66)))) *(RING(x78, x79, x80, x87, x82, x83)_>=_IF_1(x78, x79, x80, x87, x82, x83, empty(fstsplit(x83, x78)))) *(RING(x96, x97, x98, x105, x106, x101)_>=_IF_1(x96, x97, x98, x105, x106, x101, empty(fstsplit(x101, x96)))) *(RING(x114, x121, x122, x123, x118, x119)_>=_IF_1(x114, x121, x122, x123, x118, x119, empty(fstsplit(x119, x114)))) *IF_1(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) *(IF_1(cons(x1872, x1871), x141, x142, x143, x144, s(x1873), false)_>=_RING(sndsplit(s(x1873), cons(x1872, x1871)), cons(fstsplit(s(x1873), cons(x1872, x1871)), x141), x142, x143, x144, s(x1873))) *RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) *(RING(x256, x257, x252, x253, x254, x255)_>=_IF_2(x256, x257, x252, x253, x254, x255, leq(x255, length(x252)))) *(RING(x274, x275, x282, x283, x278, x279)_>=_IF_2(x274, x275, x282, x283, x278, x279, leq(x279, length(x282)))) *(RING(x286, x293, x288, x289, x290, x291)_>=_IF_2(x286, x293, x288, x289, x290, x291, leq(x291, length(x288)))) *(RING(x310, x311, x312, x313, x320, x315)_>=_IF_2(x310, x311, x312, x313, x320, x315, leq(x315, length(x312)))) *(RING(x322, x323, x324, x331, x326, x327)_>=_IF_2(x322, x323, x324, x331, x326, x327, leq(x327, length(x324)))) *(RING(x340, x341, x342, x349, x350, x345)_>=_IF_2(x340, x341, x342, x349, x350, x345, leq(x345, length(x342)))) *(RING(x358, x365, x366, x367, x362, x363)_>=_IF_2(x358, x365, x366, x367, x362, x363, leq(x363, length(x366)))) *IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) *(IF_2(x396, x397, x398, x399, x400, 0, true)_>=_IF_3(x396, x397, x398, x399, x400, 0, empty(fstsplit(0, x398)))) *(IF_2(x396, x397, x1898, x399, x400, x1892, true)_>=_IF_3(x396, x397, x1898, x399, x400, x1892, empty(fstsplit(x1892, x1898))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) *IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) *(IF_3(x506, x507, cons(x1917, x1916), x509, x510, s(x1918), false)_>=_RING(x506, x507, sndsplit(s(x1918), cons(x1917, x1916)), cons(fstsplit(s(x1918), cons(x1917, x1916)), x509), x510, s(x1918))) *IF_5(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, tail(in_2), st_2, in_3, st_3, m) *(IF_5(x676, cons(nil, x678), x679, x680, x681, x682, true)_>=_RING(x676, tail(cons(nil, x678)), x679, x680, x681, x682)) *RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) *(RING(x708, x709, x704, x705, x706, x707)_>=_IF_6(x708, x709, x704, x705, x706, x707, leq(x707, length(x706)))) *(RING(x726, x727, x734, x735, x730, x731)_>=_IF_6(x726, x727, x734, x735, x730, x731, leq(x731, length(x730)))) *(RING(x738, x745, x740, x741, x742, x743)_>=_IF_6(x738, x745, x740, x741, x742, x743, leq(x743, length(x742)))) *(RING(x762, x763, x764, x765, x772, x767)_>=_IF_6(x762, x763, x764, x765, x772, x767, leq(x767, length(x772)))) *(RING(x774, x775, x776, x783, x778, x779)_>=_IF_6(x774, x775, x776, x783, x778, x779, leq(x779, length(x778)))) *(RING(x792, x793, x794, x801, x802, x797)_>=_IF_6(x792, x793, x794, x801, x802, x797, leq(x797, length(x802)))) *(RING(x810, x817, x818, x819, x814, x815)_>=_IF_6(x810, x817, x818, x819, x814, x815, leq(x815, length(x814)))) *IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) *(IF_6(x872, x873, x874, x875, x876, 0, true)_>=_IF_7(x872, x873, x874, x875, x876, 0, empty(fstsplit(0, x876)))) *(IF_6(x872, x873, x874, x875, x1963, x1957, true)_>=_IF_7(x872, x873, x874, x875, x1963, x1957, empty(fstsplit(x1957, x1963))) ==> IF_6(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), true)_>=_IF_7(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), empty(fstsplit(s(x1957), cons(x1964, x1963))))) *IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) *(IF_7(x982, x983, x984, x985, cons(x1982, x1981), s(x1983), false)_>=_RING(x982, x983, x984, x985, sndsplit(s(x1983), cons(x1982, x1981)), s(x1983))) *IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) *(IF_9(x1135, x1136, x1137, cons(nil, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(nil, x1139)), x1140, x1141)) *IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) *(IF_6(x1184, x1185, x1186, x1187, nil, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, nil, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), nil))))) *(IF_6(x1184, x1185, x1186, x1187, x2022, x2014, false)_>=_IF_8(x1184, x1185, x1186, x1187, x2022, x2014, empty(fstsplit(x2014, app(map_f(three, head(x1187)), x2022)))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) *IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) *(IF_8(x1312, x1313, x1314, x1315, x1316, s(x2046), false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(s(x2046), app(map_f(three, head(x1315)), x1316)), s(x2046))) *IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) *(IF_2(x1368, x1369, nil, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, nil, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), nil))))) *(IF_2(x1368, x1369, x2066, x1371, x1372, x2058, false)_>=_IF_4(x1368, x1369, x2066, x1371, x1372, x2058, empty(fstsplit(x2058, app(map_f(two, head(x1369)), x2066)))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) *IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) *(IF_4(x1532, x1533, x1534, x1535, x1536, s(x2090), false)_>=_RING(x1532, tail(x1533), sndsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), cons(fstsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), x1535), x1536, s(x2090))) *RING(y0, cons(x0, x1), y2, y3, y4, y5) -> IF_5(y0, cons(x0, x1), y2, y3, y4, y5, empty(map_f(two, x0))) *(RING(x1576, cons(x1577, x1571), x1572, x1573, x1574, x1575)_>=_IF_5(x1576, cons(x1577, x1571), x1572, x1573, x1574, x1575, empty(map_f(two, x1577)))) *(RING(x1595, cons(x1602, x1603), x1604, x1605, x1599, x1600)_>=_IF_5(x1595, cons(x1602, x1603), x1604, x1605, x1599, x1600, empty(map_f(two, x1602)))) *(RING(x1608, cons(x1615, x1616), x1610, x1611, x1612, x1613)_>=_IF_5(x1608, cons(x1615, x1616), x1610, x1611, x1612, x1613, empty(map_f(two, x1615)))) *(RING(x1633, cons(x1640, x1641), x1635, x1636, x1644, x1638)_>=_IF_5(x1633, cons(x1640, x1641), x1635, x1636, x1644, x1638, empty(map_f(two, x1640)))) *(RING(x1646, cons(x1653, x1654), x1648, x1656, x1650, x1651)_>=_IF_5(x1646, cons(x1653, x1654), x1648, x1656, x1650, x1651, empty(map_f(two, x1653)))) *(RING(x1665, cons(x1672, x1673), x1667, x1675, x1676, x1670)_>=_IF_5(x1665, cons(x1672, x1673), x1667, x1675, x1676, x1670, empty(map_f(two, x1672)))) *(RING(x1684, cons(x1691, x1692), x1693, x1694, x1688, x1689)_>=_IF_5(x1684, cons(x1691, x1692), x1693, x1694, x1688, x1689, empty(map_f(two, x1691)))) *RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) *(RING(x1723, x1724, x1719, cons(x1726, x1727), x1721, x1722)_>=_IF_9(x1723, x1724, x1719, cons(x1726, x1727), x1721, x1722, empty(map_f(three, x1726)))) *(RING(x1742, x1743, x1750, cons(x1751, x1745), x1746, x1747)_>=_IF_9(x1742, x1743, x1750, cons(x1751, x1745), x1746, x1747, empty(map_f(three, x1751)))) *(RING(x1755, x1762, x1757, cons(x1764, x1765), x1759, x1760)_>=_IF_9(x1755, x1762, x1757, cons(x1764, x1765), x1759, x1760, empty(map_f(three, x1764)))) *(RING(x1780, x1781, x1782, cons(x1789, x1790), x1791, x1785)_>=_IF_9(x1780, x1781, x1782, cons(x1789, x1790), x1791, x1785, empty(map_f(three, x1789)))) *(RING(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798)_>=_IF_9(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798, empty(map_f(three, x1802)))) *(RING(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817)_>=_IF_9(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817, empty(map_f(three, x1821)))) *(RING(x1831, x1838, x1839, cons(x1840, x1834), x1835, x1836)_>=_IF_9(x1831, x1838, x1839, cons(x1840, x1834), x1835, x1836, empty(map_f(three, x1840)))) The constraints for P_> respective P_bound are constructed from P__>=_ where we just replace every occurence of "t _>=_ s" in P__>=_ by "t > s" respective "t _>=_ c". Here c stands for the fresh constant used for P_bound. Using the following integer polynomial ordering the resulting constraints can be solved Polynomial interpretation [NONINF]: POL(0) = 0 POL(IF_1(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = -1 - x_1 - x_6 - x_7 POL(IF_2(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = -1 - x_1 - x_6 + x_7 POL(IF_3(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = -1 - x_1 - x_6 - x_7 POL(IF_4(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = -1 - x_1 - x_6 - x_7 POL(IF_5(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = -1 - x_1 - x_2 - x_6 - x_7 POL(IF_6(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = -1 - x_1 - x_6 + x_7 POL(IF_7(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = -1 - x_1 - x_6 - x_7 POL(IF_8(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = -1 - x_1 - x_6 - x_7 POL(IF_9(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = -1 - x_1 - x_4 - x_6 - x_7 POL(RING(x_1, x_2, x_3, x_4, x_5, x_6)) = -1 - x_1 - x_6 POL(app(x_1, x_2)) = 0 POL(c) = -2 POL(cons(x_1, x_2)) = 0 POL(empty(x_1)) = 0 POL(f(x_1, x_2)) = x_1 + x_2 POL(false) = 0 POL(fstsplit(x_1, x_2)) = 0 POL(head(x_1)) = 0 POL(length(x_1)) = 0 POL(leq(x_1, x_2)) = 0 POL(map_f(x_1, x_2)) = 0 POL(nil) = 1 POL(s(x_1)) = 1 POL(sndsplit(x_1, x_2)) = x_1 POL(tail(x_1)) = 0 POL(three) = 0 POL(true) = 0 POL(two) = 0 The following pairs are in P_>: IF_1(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) The following pairs are in P_bound: IF_1(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m) The following rules are usable: true -> empty(nil) false -> empty(cons(h, t)) x -> sndsplit(0, x) nil -> sndsplit(s(n), nil) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) ---------------------------------------- (61) Obligation: Q DP problem: The TRS P consists of the following rules: RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1))) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) IF_5(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, tail(in_2), st_2, in_3, st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) RING(y0, cons(x0, x1), y2, y3, y4, y5) -> IF_5(y0, cons(x0, x1), y2, y3, y4, y5, empty(map_f(two, x0))) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (62) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (63) Obligation: Q DP problem: The TRS P consists of the following rules: RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) RING(y0, cons(x0, x1), y2, y3, y4, y5) -> IF_5(y0, cons(x0, x1), y2, y3, y4, y5, empty(map_f(two, x0))) IF_5(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, tail(in_2), st_2, in_3, st_3, m) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (64) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. RING(y0, cons(x0, x1), y2, y3, y4, y5) -> IF_5(y0, cons(x0, x1), y2, y3, y4, y5, empty(map_f(two, x0))) IF_5(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, tail(in_2), st_2, in_3, st_3, m) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO,RATPOLO]: POL(0) = 0 POL(IF_2(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = [4]x_2 POL(IF_3(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = [4]x_2 POL(IF_4(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = [4]x_2 POL(IF_5(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = [1/4] + [2]x_2 POL(IF_6(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = [4]x_2 POL(IF_7(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = [4]x_2 POL(IF_8(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = [4]x_2 POL(IF_9(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = [4]x_2 POL(RING(x_1, x_2, x_3, x_4, x_5, x_6)) = [4]x_2 POL(app(x_1, x_2)) = 0 POL(cons(x_1, x_2)) = [1/4] + x_1 + [2]x_2 POL(empty(x_1)) = 0 POL(f(x_1, x_2)) = 0 POL(false) = 0 POL(fstsplit(x_1, x_2)) = 0 POL(head(x_1)) = 0 POL(length(x_1)) = 0 POL(leq(x_1, x_2)) = 0 POL(map_f(x_1, x_2)) = 0 POL(nil) = 0 POL(s(x_1)) = 0 POL(sndsplit(x_1, x_2)) = 0 POL(tail(x_1)) = [1/2]x_1 POL(three) = 0 POL(true) = 0 POL(two) = 0 The value of delta used in the strict ordering is 1/4. The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: tail(cons(h, t)) -> t ---------------------------------------- (65) Obligation: Q DP problem: The TRS P consists of the following rules: RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (66) NonInfProof (EQUIVALENT) The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: Note that final constraints are written in bold face. For Pair RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) the following chains were created: *We consider the chain IF_3(x274, x275, x276, x277, x278, x279, false) -> RING(x274, x275, sndsplit(x279, x276), cons(fstsplit(x279, x276), x277), x278, x279), RING(x280, x281, x282, x283, x284, x285) -> IF_2(x280, x281, x282, x283, x284, x285, leq(x285, length(x282))) which results in the following constraint: (1) (RING(x274, x275, sndsplit(x279, x276), cons(fstsplit(x279, x276), x277), x278, x279)=RING(x280, x281, x282, x283, x284, x285) ==> RING(x280, x281, x282, x283, x284, x285)_>=_IF_2(x280, x281, x282, x283, x284, x285, leq(x285, length(x282)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x274, x275, x282, x283, x278, x279)_>=_IF_2(x274, x275, x282, x283, x278, x279, leq(x279, length(x282)))) *We consider the chain IF_7(x310, x311, x312, x313, x314, x315, false) -> RING(x310, x311, x312, x313, sndsplit(x315, x314), x315), RING(x316, x317, x318, x319, x320, x321) -> IF_2(x316, x317, x318, x319, x320, x321, leq(x321, length(x318))) which results in the following constraint: (1) (RING(x310, x311, x312, x313, sndsplit(x315, x314), x315)=RING(x316, x317, x318, x319, x320, x321) ==> RING(x316, x317, x318, x319, x320, x321)_>=_IF_2(x316, x317, x318, x319, x320, x321, leq(x321, length(x318)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x310, x311, x312, x313, x320, x315)_>=_IF_2(x310, x311, x312, x313, x320, x315, leq(x315, length(x312)))) *We consider the chain IF_9(x322, x323, x324, x325, x326, x327, true) -> RING(x322, x323, x324, tail(x325), x326, x327), RING(x328, x329, x330, x331, x332, x333) -> IF_2(x328, x329, x330, x331, x332, x333, leq(x333, length(x330))) which results in the following constraint: (1) (RING(x322, x323, x324, tail(x325), x326, x327)=RING(x328, x329, x330, x331, x332, x333) ==> RING(x328, x329, x330, x331, x332, x333)_>=_IF_2(x328, x329, x330, x331, x332, x333, leq(x333, length(x330)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x322, x323, x324, x331, x326, x327)_>=_IF_2(x322, x323, x324, x331, x326, x327, leq(x327, length(x324)))) *We consider the chain IF_8(x340, x341, x342, x343, x344, x345, false) -> RING(x340, x341, x342, tail(x343), sndsplit(x345, app(map_f(three, head(x343)), x344)), x345), RING(x346, x347, x348, x349, x350, x351) -> IF_2(x346, x347, x348, x349, x350, x351, leq(x351, length(x348))) which results in the following constraint: (1) (RING(x340, x341, x342, tail(x343), sndsplit(x345, app(map_f(three, head(x343)), x344)), x345)=RING(x346, x347, x348, x349, x350, x351) ==> RING(x346, x347, x348, x349, x350, x351)_>=_IF_2(x346, x347, x348, x349, x350, x351, leq(x351, length(x348)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x340, x341, x342, x349, x350, x345)_>=_IF_2(x340, x341, x342, x349, x350, x345, leq(x345, length(x342)))) *We consider the chain IF_4(x358, x359, x360, x361, x362, x363, false) -> RING(x358, tail(x359), sndsplit(x363, app(map_f(two, head(x359)), x360)), cons(fstsplit(x363, app(map_f(two, head(x359)), x360)), x361), x362, x363), RING(x364, x365, x366, x367, x368, x369) -> IF_2(x364, x365, x366, x367, x368, x369, leq(x369, length(x366))) which results in the following constraint: (1) (RING(x358, tail(x359), sndsplit(x363, app(map_f(two, head(x359)), x360)), cons(fstsplit(x363, app(map_f(two, head(x359)), x360)), x361), x362, x363)=RING(x364, x365, x366, x367, x368, x369) ==> RING(x364, x365, x366, x367, x368, x369)_>=_IF_2(x364, x365, x366, x367, x368, x369, leq(x369, length(x366)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x358, x365, x366, x367, x362, x363)_>=_IF_2(x358, x365, x366, x367, x362, x363, leq(x363, length(x366)))) For Pair IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) the following chains were created: *We consider the chain RING(x396, x397, x398, x399, x400, x401) -> IF_2(x396, x397, x398, x399, x400, x401, leq(x401, length(x398))), IF_2(x402, x403, x404, x405, x406, x407, true) -> IF_3(x402, x403, x404, x405, x406, x407, empty(fstsplit(x407, x404))) which results in the following constraint: (1) (IF_2(x396, x397, x398, x399, x400, x401, leq(x401, length(x398)))=IF_2(x402, x403, x404, x405, x406, x407, true) ==> IF_2(x402, x403, x404, x405, x406, x407, true)_>=_IF_3(x402, x403, x404, x405, x406, x407, empty(fstsplit(x407, x404)))) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (length(x398)=x1888 & leq(x401, x1888)=true ==> IF_2(x396, x397, x398, x399, x400, x401, true)_>=_IF_3(x396, x397, x398, x399, x400, x401, empty(fstsplit(x401, x398)))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on leq(x401, x1888)=true which results in the following new constraints: (3) (true=true & length(x398)=x1889 ==> IF_2(x396, x397, x398, x399, x400, 0, true)_>=_IF_3(x396, x397, x398, x399, x400, 0, empty(fstsplit(0, x398)))) (4) (leq(x1892, x1891)=true & length(x398)=s(x1891) & (\/x1893,x1894,x1895,x1896,x1897:leq(x1892, x1891)=true & length(x1893)=x1891 ==> IF_2(x1894, x1895, x1893, x1896, x1897, x1892, true)_>=_IF_3(x1894, x1895, x1893, x1896, x1897, x1892, empty(fstsplit(x1892, x1893)))) ==> IF_2(x396, x397, x398, x399, x400, s(x1892), true)_>=_IF_3(x396, x397, x398, x399, x400, s(x1892), empty(fstsplit(s(x1892), x398)))) We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint: (5) (IF_2(x396, x397, x398, x399, x400, 0, true)_>=_IF_3(x396, x397, x398, x399, x400, 0, empty(fstsplit(0, x398)))) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on length(x398)=s(x1891) which results in the following new constraint: (6) (s(length(x1898))=s(x1891) & leq(x1892, x1891)=true & (\/x1893,x1894,x1895,x1896,x1897:leq(x1892, x1891)=true & length(x1893)=x1891 ==> IF_2(x1894, x1895, x1893, x1896, x1897, x1892, true)_>=_IF_3(x1894, x1895, x1893, x1896, x1897, x1892, empty(fstsplit(x1892, x1893)))) & (\/x1900,x1901,x1902,x1903,x1904,x1905,x1906,x1907,x1908,x1909,x1910:length(x1898)=s(x1900) & leq(x1901, x1900)=true & (\/x1902,x1903,x1904,x1905,x1906:leq(x1901, x1900)=true & length(x1902)=x1900 ==> IF_2(x1903, x1904, x1902, x1905, x1906, x1901, true)_>=_IF_3(x1903, x1904, x1902, x1905, x1906, x1901, empty(fstsplit(x1901, x1902)))) ==> IF_2(x1907, x1908, x1898, x1909, x1910, s(x1901), true)_>=_IF_3(x1907, x1908, x1898, x1909, x1910, s(x1901), empty(fstsplit(s(x1901), x1898)))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) We simplified constraint (6) using rules (I), (II) which results in the following new constraint: (7) (length(x1898)=x1891 & leq(x1892, x1891)=true & (\/x1893,x1894,x1895,x1896,x1897:leq(x1892, x1891)=true & length(x1893)=x1891 ==> IF_2(x1894, x1895, x1893, x1896, x1897, x1892, true)_>=_IF_3(x1894, x1895, x1893, x1896, x1897, x1892, empty(fstsplit(x1892, x1893)))) & (\/x1900,x1901,x1902,x1903,x1904,x1905,x1906,x1907,x1908,x1909,x1910:length(x1898)=s(x1900) & leq(x1901, x1900)=true & (\/x1902,x1903,x1904,x1905,x1906:leq(x1901, x1900)=true & length(x1902)=x1900 ==> IF_2(x1903, x1904, x1902, x1905, x1906, x1901, true)_>=_IF_3(x1903, x1904, x1902, x1905, x1906, x1901, empty(fstsplit(x1901, x1902)))) ==> IF_2(x1907, x1908, x1898, x1909, x1910, s(x1901), true)_>=_IF_3(x1907, x1908, x1898, x1909, x1910, s(x1901), empty(fstsplit(s(x1901), x1898)))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) We simplified constraint (7) using rule (VI) where we applied the induction hypothesis (\/x1893,x1894,x1895,x1896,x1897:leq(x1892, x1891)=true & length(x1893)=x1891 ==> IF_2(x1894, x1895, x1893, x1896, x1897, x1892, true)_>=_IF_3(x1894, x1895, x1893, x1896, x1897, x1892, empty(fstsplit(x1892, x1893)))) with sigma = [x1893 / x1898, x1894 / x396, x1895 / x397, x1896 / x399, x1897 / x400] which results in the following new constraint: (8) (IF_2(x396, x397, x1898, x399, x400, x1892, true)_>=_IF_3(x396, x397, x1898, x399, x400, x1892, empty(fstsplit(x1892, x1898))) & (\/x1900,x1901,x1902,x1903,x1904,x1905,x1906,x1907,x1908,x1909,x1910:length(x1898)=s(x1900) & leq(x1901, x1900)=true & (\/x1902,x1903,x1904,x1905,x1906:leq(x1901, x1900)=true & length(x1902)=x1900 ==> IF_2(x1903, x1904, x1902, x1905, x1906, x1901, true)_>=_IF_3(x1903, x1904, x1902, x1905, x1906, x1901, empty(fstsplit(x1901, x1902)))) ==> IF_2(x1907, x1908, x1898, x1909, x1910, s(x1901), true)_>=_IF_3(x1907, x1908, x1898, x1909, x1910, s(x1901), empty(fstsplit(s(x1901), x1898)))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) We simplified constraint (8) using rule (IV) which results in the following new constraint: (9) (IF_2(x396, x397, x1898, x399, x400, x1892, true)_>=_IF_3(x396, x397, x1898, x399, x400, x1892, empty(fstsplit(x1892, x1898))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) For Pair IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) the following chains were created: *We consider the chain IF_2(x506, x507, x508, x509, x510, x511, true) -> IF_3(x506, x507, x508, x509, x510, x511, empty(fstsplit(x511, x508))), IF_3(x512, x513, x514, x515, x516, x517, false) -> RING(x512, x513, sndsplit(x517, x514), cons(fstsplit(x517, x514), x515), x516, x517) which results in the following constraint: (1) (IF_3(x506, x507, x508, x509, x510, x511, empty(fstsplit(x511, x508)))=IF_3(x512, x513, x514, x515, x516, x517, false) ==> IF_3(x512, x513, x514, x515, x516, x517, false)_>=_RING(x512, x513, sndsplit(x517, x514), cons(fstsplit(x517, x514), x515), x516, x517)) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (fstsplit(x511, x508)=x1911 & empty(x1911)=false ==> IF_3(x506, x507, x508, x509, x510, x511, false)_>=_RING(x506, x507, sndsplit(x511, x508), cons(fstsplit(x511, x508), x509), x510, x511)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x1911)=false which results in the following new constraint: (3) (false=false & fstsplit(x511, x508)=cons(x1913, x1912) ==> IF_3(x506, x507, x508, x509, x510, x511, false)_>=_RING(x506, x507, sndsplit(x511, x508), cons(fstsplit(x511, x508), x509), x510, x511)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (fstsplit(x511, x508)=cons(x1913, x1912) ==> IF_3(x506, x507, x508, x509, x510, x511, false)_>=_RING(x506, x507, sndsplit(x511, x508), cons(fstsplit(x511, x508), x509), x510, x511)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on fstsplit(x511, x508)=cons(x1913, x1912) which results in the following new constraint: (5) (cons(x1917, fstsplit(x1918, x1916))=cons(x1913, x1912) & (\/x1919,x1920,x1921,x1922,x1923,x1924:fstsplit(x1918, x1916)=cons(x1919, x1920) ==> IF_3(x1921, x1922, x1916, x1923, x1924, x1918, false)_>=_RING(x1921, x1922, sndsplit(x1918, x1916), cons(fstsplit(x1918, x1916), x1923), x1924, x1918)) ==> IF_3(x506, x507, cons(x1917, x1916), x509, x510, s(x1918), false)_>=_RING(x506, x507, sndsplit(s(x1918), cons(x1917, x1916)), cons(fstsplit(s(x1918), cons(x1917, x1916)), x509), x510, s(x1918))) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (6) (IF_3(x506, x507, cons(x1917, x1916), x509, x510, s(x1918), false)_>=_RING(x506, x507, sndsplit(s(x1918), cons(x1917, x1916)), cons(fstsplit(s(x1918), cons(x1917, x1916)), x509), x510, s(x1918))) For Pair RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) the following chains were created: *We consider the chain IF_3(x726, x727, x728, x729, x730, x731, false) -> RING(x726, x727, sndsplit(x731, x728), cons(fstsplit(x731, x728), x729), x730, x731), RING(x732, x733, x734, x735, x736, x737) -> IF_6(x732, x733, x734, x735, x736, x737, leq(x737, length(x736))) which results in the following constraint: (1) (RING(x726, x727, sndsplit(x731, x728), cons(fstsplit(x731, x728), x729), x730, x731)=RING(x732, x733, x734, x735, x736, x737) ==> RING(x732, x733, x734, x735, x736, x737)_>=_IF_6(x732, x733, x734, x735, x736, x737, leq(x737, length(x736)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x726, x727, x734, x735, x730, x731)_>=_IF_6(x726, x727, x734, x735, x730, x731, leq(x731, length(x730)))) *We consider the chain IF_7(x762, x763, x764, x765, x766, x767, false) -> RING(x762, x763, x764, x765, sndsplit(x767, x766), x767), RING(x768, x769, x770, x771, x772, x773) -> IF_6(x768, x769, x770, x771, x772, x773, leq(x773, length(x772))) which results in the following constraint: (1) (RING(x762, x763, x764, x765, sndsplit(x767, x766), x767)=RING(x768, x769, x770, x771, x772, x773) ==> RING(x768, x769, x770, x771, x772, x773)_>=_IF_6(x768, x769, x770, x771, x772, x773, leq(x773, length(x772)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x762, x763, x764, x765, x772, x767)_>=_IF_6(x762, x763, x764, x765, x772, x767, leq(x767, length(x772)))) *We consider the chain IF_9(x774, x775, x776, x777, x778, x779, true) -> RING(x774, x775, x776, tail(x777), x778, x779), RING(x780, x781, x782, x783, x784, x785) -> IF_6(x780, x781, x782, x783, x784, x785, leq(x785, length(x784))) which results in the following constraint: (1) (RING(x774, x775, x776, tail(x777), x778, x779)=RING(x780, x781, x782, x783, x784, x785) ==> RING(x780, x781, x782, x783, x784, x785)_>=_IF_6(x780, x781, x782, x783, x784, x785, leq(x785, length(x784)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x774, x775, x776, x783, x778, x779)_>=_IF_6(x774, x775, x776, x783, x778, x779, leq(x779, length(x778)))) *We consider the chain IF_8(x792, x793, x794, x795, x796, x797, false) -> RING(x792, x793, x794, tail(x795), sndsplit(x797, app(map_f(three, head(x795)), x796)), x797), RING(x798, x799, x800, x801, x802, x803) -> IF_6(x798, x799, x800, x801, x802, x803, leq(x803, length(x802))) which results in the following constraint: (1) (RING(x792, x793, x794, tail(x795), sndsplit(x797, app(map_f(three, head(x795)), x796)), x797)=RING(x798, x799, x800, x801, x802, x803) ==> RING(x798, x799, x800, x801, x802, x803)_>=_IF_6(x798, x799, x800, x801, x802, x803, leq(x803, length(x802)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x792, x793, x794, x801, x802, x797)_>=_IF_6(x792, x793, x794, x801, x802, x797, leq(x797, length(x802)))) *We consider the chain IF_4(x810, x811, x812, x813, x814, x815, false) -> RING(x810, tail(x811), sndsplit(x815, app(map_f(two, head(x811)), x812)), cons(fstsplit(x815, app(map_f(two, head(x811)), x812)), x813), x814, x815), RING(x816, x817, x818, x819, x820, x821) -> IF_6(x816, x817, x818, x819, x820, x821, leq(x821, length(x820))) which results in the following constraint: (1) (RING(x810, tail(x811), sndsplit(x815, app(map_f(two, head(x811)), x812)), cons(fstsplit(x815, app(map_f(two, head(x811)), x812)), x813), x814, x815)=RING(x816, x817, x818, x819, x820, x821) ==> RING(x816, x817, x818, x819, x820, x821)_>=_IF_6(x816, x817, x818, x819, x820, x821, leq(x821, length(x820)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x810, x817, x818, x819, x814, x815)_>=_IF_6(x810, x817, x818, x819, x814, x815, leq(x815, length(x814)))) For Pair IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) the following chains were created: *We consider the chain RING(x872, x873, x874, x875, x876, x877) -> IF_6(x872, x873, x874, x875, x876, x877, leq(x877, length(x876))), IF_6(x878, x879, x880, x881, x882, x883, true) -> IF_7(x878, x879, x880, x881, x882, x883, empty(fstsplit(x883, x882))) which results in the following constraint: (1) (IF_6(x872, x873, x874, x875, x876, x877, leq(x877, length(x876)))=IF_6(x878, x879, x880, x881, x882, x883, true) ==> IF_6(x878, x879, x880, x881, x882, x883, true)_>=_IF_7(x878, x879, x880, x881, x882, x883, empty(fstsplit(x883, x882)))) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (length(x876)=x1953 & leq(x877, x1953)=true ==> IF_6(x872, x873, x874, x875, x876, x877, true)_>=_IF_7(x872, x873, x874, x875, x876, x877, empty(fstsplit(x877, x876)))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on leq(x877, x1953)=true which results in the following new constraints: (3) (true=true & length(x876)=x1954 ==> IF_6(x872, x873, x874, x875, x876, 0, true)_>=_IF_7(x872, x873, x874, x875, x876, 0, empty(fstsplit(0, x876)))) (4) (leq(x1957, x1956)=true & length(x876)=s(x1956) & (\/x1958,x1959,x1960,x1961,x1962:leq(x1957, x1956)=true & length(x1958)=x1956 ==> IF_6(x1959, x1960, x1961, x1962, x1958, x1957, true)_>=_IF_7(x1959, x1960, x1961, x1962, x1958, x1957, empty(fstsplit(x1957, x1958)))) ==> IF_6(x872, x873, x874, x875, x876, s(x1957), true)_>=_IF_7(x872, x873, x874, x875, x876, s(x1957), empty(fstsplit(s(x1957), x876)))) We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint: (5) (IF_6(x872, x873, x874, x875, x876, 0, true)_>=_IF_7(x872, x873, x874, x875, x876, 0, empty(fstsplit(0, x876)))) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on length(x876)=s(x1956) which results in the following new constraint: (6) (s(length(x1963))=s(x1956) & leq(x1957, x1956)=true & (\/x1958,x1959,x1960,x1961,x1962:leq(x1957, x1956)=true & length(x1958)=x1956 ==> IF_6(x1959, x1960, x1961, x1962, x1958, x1957, true)_>=_IF_7(x1959, x1960, x1961, x1962, x1958, x1957, empty(fstsplit(x1957, x1958)))) & (\/x1965,x1966,x1967,x1968,x1969,x1970,x1971,x1972,x1973,x1974,x1975:length(x1963)=s(x1965) & leq(x1966, x1965)=true & (\/x1967,x1968,x1969,x1970,x1971:leq(x1966, x1965)=true & length(x1967)=x1965 ==> IF_6(x1968, x1969, x1970, x1971, x1967, x1966, true)_>=_IF_7(x1968, x1969, x1970, x1971, x1967, x1966, empty(fstsplit(x1966, x1967)))) ==> IF_6(x1972, x1973, x1974, x1975, x1963, s(x1966), true)_>=_IF_7(x1972, x1973, x1974, x1975, x1963, s(x1966), empty(fstsplit(s(x1966), x1963)))) ==> IF_6(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), true)_>=_IF_7(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), empty(fstsplit(s(x1957), cons(x1964, x1963))))) We simplified constraint (6) using rules (I), (II) which results in the following new constraint: (7) (length(x1963)=x1956 & leq(x1957, x1956)=true & (\/x1958,x1959,x1960,x1961,x1962:leq(x1957, x1956)=true & length(x1958)=x1956 ==> IF_6(x1959, x1960, x1961, x1962, x1958, x1957, true)_>=_IF_7(x1959, x1960, x1961, x1962, x1958, x1957, empty(fstsplit(x1957, x1958)))) & (\/x1965,x1966,x1967,x1968,x1969,x1970,x1971,x1972,x1973,x1974,x1975:length(x1963)=s(x1965) & leq(x1966, x1965)=true & (\/x1967,x1968,x1969,x1970,x1971:leq(x1966, x1965)=true & length(x1967)=x1965 ==> IF_6(x1968, x1969, x1970, x1971, x1967, x1966, true)_>=_IF_7(x1968, x1969, x1970, x1971, x1967, x1966, empty(fstsplit(x1966, x1967)))) ==> IF_6(x1972, x1973, x1974, x1975, x1963, s(x1966), true)_>=_IF_7(x1972, x1973, x1974, x1975, x1963, s(x1966), empty(fstsplit(s(x1966), x1963)))) ==> IF_6(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), true)_>=_IF_7(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), empty(fstsplit(s(x1957), cons(x1964, x1963))))) We simplified constraint (7) using rule (VI) where we applied the induction hypothesis (\/x1958,x1959,x1960,x1961,x1962:leq(x1957, x1956)=true & length(x1958)=x1956 ==> IF_6(x1959, x1960, x1961, x1962, x1958, x1957, true)_>=_IF_7(x1959, x1960, x1961, x1962, x1958, x1957, empty(fstsplit(x1957, x1958)))) with sigma = [x1958 / x1963, x1959 / x872, x1960 / x873, x1961 / x874, x1962 / x875] which results in the following new constraint: (8) (IF_6(x872, x873, x874, x875, x1963, x1957, true)_>=_IF_7(x872, x873, x874, x875, x1963, x1957, empty(fstsplit(x1957, x1963))) & (\/x1965,x1966,x1967,x1968,x1969,x1970,x1971,x1972,x1973,x1974,x1975:length(x1963)=s(x1965) & leq(x1966, x1965)=true & (\/x1967,x1968,x1969,x1970,x1971:leq(x1966, x1965)=true & length(x1967)=x1965 ==> IF_6(x1968, x1969, x1970, x1971, x1967, x1966, true)_>=_IF_7(x1968, x1969, x1970, x1971, x1967, x1966, empty(fstsplit(x1966, x1967)))) ==> IF_6(x1972, x1973, x1974, x1975, x1963, s(x1966), true)_>=_IF_7(x1972, x1973, x1974, x1975, x1963, s(x1966), empty(fstsplit(s(x1966), x1963)))) ==> IF_6(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), true)_>=_IF_7(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), empty(fstsplit(s(x1957), cons(x1964, x1963))))) We simplified constraint (8) using rule (IV) which results in the following new constraint: (9) (IF_6(x872, x873, x874, x875, x1963, x1957, true)_>=_IF_7(x872, x873, x874, x875, x1963, x1957, empty(fstsplit(x1957, x1963))) ==> IF_6(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), true)_>=_IF_7(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), empty(fstsplit(s(x1957), cons(x1964, x1963))))) For Pair IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) the following chains were created: *We consider the chain IF_6(x982, x983, x984, x985, x986, x987, true) -> IF_7(x982, x983, x984, x985, x986, x987, empty(fstsplit(x987, x986))), IF_7(x988, x989, x990, x991, x992, x993, false) -> RING(x988, x989, x990, x991, sndsplit(x993, x992), x993) which results in the following constraint: (1) (IF_7(x982, x983, x984, x985, x986, x987, empty(fstsplit(x987, x986)))=IF_7(x988, x989, x990, x991, x992, x993, false) ==> IF_7(x988, x989, x990, x991, x992, x993, false)_>=_RING(x988, x989, x990, x991, sndsplit(x993, x992), x993)) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (fstsplit(x987, x986)=x1976 & empty(x1976)=false ==> IF_7(x982, x983, x984, x985, x986, x987, false)_>=_RING(x982, x983, x984, x985, sndsplit(x987, x986), x987)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x1976)=false which results in the following new constraint: (3) (false=false & fstsplit(x987, x986)=cons(x1978, x1977) ==> IF_7(x982, x983, x984, x985, x986, x987, false)_>=_RING(x982, x983, x984, x985, sndsplit(x987, x986), x987)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (fstsplit(x987, x986)=cons(x1978, x1977) ==> IF_7(x982, x983, x984, x985, x986, x987, false)_>=_RING(x982, x983, x984, x985, sndsplit(x987, x986), x987)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on fstsplit(x987, x986)=cons(x1978, x1977) which results in the following new constraint: (5) (cons(x1982, fstsplit(x1983, x1981))=cons(x1978, x1977) & (\/x1984,x1985,x1986,x1987,x1988,x1989:fstsplit(x1983, x1981)=cons(x1984, x1985) ==> IF_7(x1986, x1987, x1988, x1989, x1981, x1983, false)_>=_RING(x1986, x1987, x1988, x1989, sndsplit(x1983, x1981), x1983)) ==> IF_7(x982, x983, x984, x985, cons(x1982, x1981), s(x1983), false)_>=_RING(x982, x983, x984, x985, sndsplit(s(x1983), cons(x1982, x1981)), s(x1983))) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (6) (IF_7(x982, x983, x984, x985, cons(x1982, x1981), s(x1983), false)_>=_RING(x982, x983, x984, x985, sndsplit(s(x1983), cons(x1982, x1981)), s(x1983))) For Pair IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) the following chains were created: *We consider the chain RING(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141) -> IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, empty(map_f(three, x1138))), IF_9(x1142, x1143, x1144, x1145, x1146, x1147, true) -> RING(x1142, x1143, x1144, tail(x1145), x1146, x1147) which results in the following constraint: (1) (IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, empty(map_f(three, x1138)))=IF_9(x1142, x1143, x1144, x1145, x1146, x1147, true) ==> IF_9(x1142, x1143, x1144, x1145, x1146, x1147, true)_>=_RING(x1142, x1143, x1144, tail(x1145), x1146, x1147)) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (three=x1991 & map_f(x1991, x1138)=x1990 & empty(x1990)=true ==> IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(x1138, x1139)), x1140, x1141)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x1990)=true which results in the following new constraint: (3) (true=true & three=x1991 & map_f(x1991, x1138)=nil ==> IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(x1138, x1139)), x1140, x1141)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (three=x1991 & map_f(x1991, x1138)=nil ==> IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(x1138, x1139)), x1140, x1141)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on map_f(x1991, x1138)=nil which results in the following new constraints: (5) (nil=nil & three=x1994 ==> IF_9(x1135, x1136, x1137, cons(nil, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(nil, x1139)), x1140, x1141)) (6) (app(f(x1997, x1996), map_f(x1997, x1995))=nil & three=x1997 & (\/x1998,x1999,x2000,x2001,x2002,x2003:map_f(x1997, x1995)=nil & three=x1997 ==> IF_9(x1998, x1999, x2000, cons(x1995, x2001), x2002, x2003, true)_>=_RING(x1998, x1999, x2000, tail(cons(x1995, x2001)), x2002, x2003)) ==> IF_9(x1135, x1136, x1137, cons(cons(x1996, x1995), x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(cons(x1996, x1995), x1139)), x1140, x1141)) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (7) (IF_9(x1135, x1136, x1137, cons(nil, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(nil, x1139)), x1140, x1141)) We simplified constraint (6) using rules (IV), (VII) which results in the following new constraint: (8) (f(x1997, x1996)=x2004 & map_f(x1997, x1995)=x2005 & app(x2004, x2005)=nil & three=x1997 ==> IF_9(x1135, x1136, x1137, cons(cons(x1996, x1995), x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(cons(x1996, x1995), x1139)), x1140, x1141)) We simplified constraint (8) using rule (V) (with possible (I) afterwards) using induction on app(x2004, x2005)=nil which results in the following new constraint: (9) (x2006=nil & f(x1997, x1996)=nil & map_f(x1997, x1995)=x2006 & three=x1997 ==> IF_9(x1135, x1136, x1137, cons(cons(x1996, x1995), x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(cons(x1996, x1995), x1139)), x1140, x1141)) We solved constraint (9) using rules (I), (II). For Pair IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) the following chains were created: *We consider the chain RING(x1184, x1185, x1186, x1187, x1188, x1189) -> IF_6(x1184, x1185, x1186, x1187, x1188, x1189, leq(x1189, length(x1188))), IF_6(x1190, x1191, x1192, x1193, x1194, x1195, false) -> IF_8(x1190, x1191, x1192, x1193, x1194, x1195, empty(fstsplit(x1195, app(map_f(three, head(x1193)), x1194)))) which results in the following constraint: (1) (IF_6(x1184, x1185, x1186, x1187, x1188, x1189, leq(x1189, length(x1188)))=IF_6(x1190, x1191, x1192, x1193, x1194, x1195, false) ==> IF_6(x1190, x1191, x1192, x1193, x1194, x1195, false)_>=_IF_8(x1190, x1191, x1192, x1193, x1194, x1195, empty(fstsplit(x1195, app(map_f(three, head(x1193)), x1194))))) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (length(x1188)=x2010 & leq(x1189, x2010)=false ==> IF_6(x1184, x1185, x1186, x1187, x1188, x1189, false)_>=_IF_8(x1184, x1185, x1186, x1187, x1188, x1189, empty(fstsplit(x1189, app(map_f(three, head(x1187)), x1188))))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on leq(x1189, x2010)=false which results in the following new constraints: (3) (false=false & length(x1188)=0 ==> IF_6(x1184, x1185, x1186, x1187, x1188, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, x1188, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), x1188))))) (4) (leq(x2014, x2013)=false & length(x1188)=s(x2013) & (\/x2015,x2016,x2017,x2018,x2019:leq(x2014, x2013)=false & length(x2015)=x2013 ==> IF_6(x2016, x2017, x2018, x2019, x2015, x2014, false)_>=_IF_8(x2016, x2017, x2018, x2019, x2015, x2014, empty(fstsplit(x2014, app(map_f(three, head(x2019)), x2015))))) ==> IF_6(x1184, x1185, x1186, x1187, x1188, s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, x1188, s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), x1188))))) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (5) (length(x1188)=0 ==> IF_6(x1184, x1185, x1186, x1187, x1188, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, x1188, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), x1188))))) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on length(x1188)=s(x2013) which results in the following new constraint: (6) (s(length(x2022))=s(x2013) & leq(x2014, x2013)=false & (\/x2015,x2016,x2017,x2018,x2019:leq(x2014, x2013)=false & length(x2015)=x2013 ==> IF_6(x2016, x2017, x2018, x2019, x2015, x2014, false)_>=_IF_8(x2016, x2017, x2018, x2019, x2015, x2014, empty(fstsplit(x2014, app(map_f(three, head(x2019)), x2015))))) & (\/x2024,x2025,x2026,x2027,x2028,x2029,x2030,x2031,x2032,x2033,x2034:length(x2022)=s(x2024) & leq(x2025, x2024)=false & (\/x2026,x2027,x2028,x2029,x2030:leq(x2025, x2024)=false & length(x2026)=x2024 ==> IF_6(x2027, x2028, x2029, x2030, x2026, x2025, false)_>=_IF_8(x2027, x2028, x2029, x2030, x2026, x2025, empty(fstsplit(x2025, app(map_f(three, head(x2030)), x2026))))) ==> IF_6(x2031, x2032, x2033, x2034, x2022, s(x2025), false)_>=_IF_8(x2031, x2032, x2033, x2034, x2022, s(x2025), empty(fstsplit(s(x2025), app(map_f(three, head(x2034)), x2022))))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on length(x1188)=0 which results in the following new constraint: (7) (0=0 ==> IF_6(x1184, x1185, x1186, x1187, nil, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, nil, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), nil))))) We simplified constraint (7) using rules (I), (II) which results in the following new constraint: (8) (IF_6(x1184, x1185, x1186, x1187, nil, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, nil, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), nil))))) We simplified constraint (6) using rules (I), (II) which results in the following new constraint: (9) (length(x2022)=x2013 & leq(x2014, x2013)=false & (\/x2015,x2016,x2017,x2018,x2019:leq(x2014, x2013)=false & length(x2015)=x2013 ==> IF_6(x2016, x2017, x2018, x2019, x2015, x2014, false)_>=_IF_8(x2016, x2017, x2018, x2019, x2015, x2014, empty(fstsplit(x2014, app(map_f(three, head(x2019)), x2015))))) & (\/x2024,x2025,x2026,x2027,x2028,x2029,x2030,x2031,x2032,x2033,x2034:length(x2022)=s(x2024) & leq(x2025, x2024)=false & (\/x2026,x2027,x2028,x2029,x2030:leq(x2025, x2024)=false & length(x2026)=x2024 ==> IF_6(x2027, x2028, x2029, x2030, x2026, x2025, false)_>=_IF_8(x2027, x2028, x2029, x2030, x2026, x2025, empty(fstsplit(x2025, app(map_f(three, head(x2030)), x2026))))) ==> IF_6(x2031, x2032, x2033, x2034, x2022, s(x2025), false)_>=_IF_8(x2031, x2032, x2033, x2034, x2022, s(x2025), empty(fstsplit(s(x2025), app(map_f(three, head(x2034)), x2022))))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) We simplified constraint (9) using rule (VI) where we applied the induction hypothesis (\/x2015,x2016,x2017,x2018,x2019:leq(x2014, x2013)=false & length(x2015)=x2013 ==> IF_6(x2016, x2017, x2018, x2019, x2015, x2014, false)_>=_IF_8(x2016, x2017, x2018, x2019, x2015, x2014, empty(fstsplit(x2014, app(map_f(three, head(x2019)), x2015))))) with sigma = [x2015 / x2022, x2016 / x1184, x2017 / x1185, x2018 / x1186, x2019 / x1187] which results in the following new constraint: (10) (IF_6(x1184, x1185, x1186, x1187, x2022, x2014, false)_>=_IF_8(x1184, x1185, x1186, x1187, x2022, x2014, empty(fstsplit(x2014, app(map_f(three, head(x1187)), x2022)))) & (\/x2024,x2025,x2026,x2027,x2028,x2029,x2030,x2031,x2032,x2033,x2034:length(x2022)=s(x2024) & leq(x2025, x2024)=false & (\/x2026,x2027,x2028,x2029,x2030:leq(x2025, x2024)=false & length(x2026)=x2024 ==> IF_6(x2027, x2028, x2029, x2030, x2026, x2025, false)_>=_IF_8(x2027, x2028, x2029, x2030, x2026, x2025, empty(fstsplit(x2025, app(map_f(three, head(x2030)), x2026))))) ==> IF_6(x2031, x2032, x2033, x2034, x2022, s(x2025), false)_>=_IF_8(x2031, x2032, x2033, x2034, x2022, s(x2025), empty(fstsplit(s(x2025), app(map_f(three, head(x2034)), x2022))))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) We simplified constraint (10) using rule (IV) which results in the following new constraint: (11) (IF_6(x1184, x1185, x1186, x1187, x2022, x2014, false)_>=_IF_8(x1184, x1185, x1186, x1187, x2022, x2014, empty(fstsplit(x2014, app(map_f(three, head(x1187)), x2022)))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) For Pair IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) the following chains were created: *We consider the chain IF_6(x1312, x1313, x1314, x1315, x1316, x1317, false) -> IF_8(x1312, x1313, x1314, x1315, x1316, x1317, empty(fstsplit(x1317, app(map_f(three, head(x1315)), x1316)))), IF_8(x1318, x1319, x1320, x1321, x1322, x1323, false) -> RING(x1318, x1319, x1320, tail(x1321), sndsplit(x1323, app(map_f(three, head(x1321)), x1322)), x1323) which results in the following constraint: (1) (IF_8(x1312, x1313, x1314, x1315, x1316, x1317, empty(fstsplit(x1317, app(map_f(three, head(x1315)), x1316))))=IF_8(x1318, x1319, x1320, x1321, x1322, x1323, false) ==> IF_8(x1318, x1319, x1320, x1321, x1322, x1323, false)_>=_RING(x1318, x1319, x1320, tail(x1321), sndsplit(x1323, app(map_f(three, head(x1321)), x1322)), x1323)) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (fstsplit(x1317, x2036)=x2035 & empty(x2035)=false ==> IF_8(x1312, x1313, x1314, x1315, x1316, x1317, false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(x1317, app(map_f(three, head(x1315)), x1316)), x1317)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x2035)=false which results in the following new constraint: (3) (false=false & fstsplit(x1317, x2036)=cons(x2041, x2040) ==> IF_8(x1312, x1313, x1314, x1315, x1316, x1317, false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(x1317, app(map_f(three, head(x1315)), x1316)), x1317)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (fstsplit(x1317, x2036)=cons(x2041, x2040) ==> IF_8(x1312, x1313, x1314, x1315, x1316, x1317, false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(x1317, app(map_f(three, head(x1315)), x1316)), x1317)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on fstsplit(x1317, x2036)=cons(x2041, x2040) which results in the following new constraint: (5) (cons(x2045, fstsplit(x2046, x2044))=cons(x2041, x2040) & (\/x2047,x2048,x2049,x2050,x2051,x2052,x2053:fstsplit(x2046, x2044)=cons(x2047, x2048) ==> IF_8(x2049, x2050, x2051, x2052, x2053, x2046, false)_>=_RING(x2049, x2050, x2051, tail(x2052), sndsplit(x2046, app(map_f(three, head(x2052)), x2053)), x2046)) ==> IF_8(x1312, x1313, x1314, x1315, x1316, s(x2046), false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(s(x2046), app(map_f(three, head(x1315)), x1316)), s(x2046))) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (6) (IF_8(x1312, x1313, x1314, x1315, x1316, s(x2046), false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(s(x2046), app(map_f(three, head(x1315)), x1316)), s(x2046))) For Pair IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) the following chains were created: *We consider the chain RING(x1368, x1369, x1370, x1371, x1372, x1373) -> IF_2(x1368, x1369, x1370, x1371, x1372, x1373, leq(x1373, length(x1370))), IF_2(x1374, x1375, x1376, x1377, x1378, x1379, false) -> IF_4(x1374, x1375, x1376, x1377, x1378, x1379, empty(fstsplit(x1379, app(map_f(two, head(x1375)), x1376)))) which results in the following constraint: (1) (IF_2(x1368, x1369, x1370, x1371, x1372, x1373, leq(x1373, length(x1370)))=IF_2(x1374, x1375, x1376, x1377, x1378, x1379, false) ==> IF_2(x1374, x1375, x1376, x1377, x1378, x1379, false)_>=_IF_4(x1374, x1375, x1376, x1377, x1378, x1379, empty(fstsplit(x1379, app(map_f(two, head(x1375)), x1376))))) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (length(x1370)=x2054 & leq(x1373, x2054)=false ==> IF_2(x1368, x1369, x1370, x1371, x1372, x1373, false)_>=_IF_4(x1368, x1369, x1370, x1371, x1372, x1373, empty(fstsplit(x1373, app(map_f(two, head(x1369)), x1370))))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on leq(x1373, x2054)=false which results in the following new constraints: (3) (false=false & length(x1370)=0 ==> IF_2(x1368, x1369, x1370, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, x1370, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), x1370))))) (4) (leq(x2058, x2057)=false & length(x1370)=s(x2057) & (\/x2059,x2060,x2061,x2062,x2063:leq(x2058, x2057)=false & length(x2059)=x2057 ==> IF_2(x2060, x2061, x2059, x2062, x2063, x2058, false)_>=_IF_4(x2060, x2061, x2059, x2062, x2063, x2058, empty(fstsplit(x2058, app(map_f(two, head(x2061)), x2059))))) ==> IF_2(x1368, x1369, x1370, x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, x1370, x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), x1370))))) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (5) (length(x1370)=0 ==> IF_2(x1368, x1369, x1370, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, x1370, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), x1370))))) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on length(x1370)=s(x2057) which results in the following new constraint: (6) (s(length(x2066))=s(x2057) & leq(x2058, x2057)=false & (\/x2059,x2060,x2061,x2062,x2063:leq(x2058, x2057)=false & length(x2059)=x2057 ==> IF_2(x2060, x2061, x2059, x2062, x2063, x2058, false)_>=_IF_4(x2060, x2061, x2059, x2062, x2063, x2058, empty(fstsplit(x2058, app(map_f(two, head(x2061)), x2059))))) & (\/x2068,x2069,x2070,x2071,x2072,x2073,x2074,x2075,x2076,x2077,x2078:length(x2066)=s(x2068) & leq(x2069, x2068)=false & (\/x2070,x2071,x2072,x2073,x2074:leq(x2069, x2068)=false & length(x2070)=x2068 ==> IF_2(x2071, x2072, x2070, x2073, x2074, x2069, false)_>=_IF_4(x2071, x2072, x2070, x2073, x2074, x2069, empty(fstsplit(x2069, app(map_f(two, head(x2072)), x2070))))) ==> IF_2(x2075, x2076, x2066, x2077, x2078, s(x2069), false)_>=_IF_4(x2075, x2076, x2066, x2077, x2078, s(x2069), empty(fstsplit(s(x2069), app(map_f(two, head(x2076)), x2066))))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on length(x1370)=0 which results in the following new constraint: (7) (0=0 ==> IF_2(x1368, x1369, nil, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, nil, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), nil))))) We simplified constraint (7) using rules (I), (II) which results in the following new constraint: (8) (IF_2(x1368, x1369, nil, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, nil, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), nil))))) We simplified constraint (6) using rules (I), (II) which results in the following new constraint: (9) (length(x2066)=x2057 & leq(x2058, x2057)=false & (\/x2059,x2060,x2061,x2062,x2063:leq(x2058, x2057)=false & length(x2059)=x2057 ==> IF_2(x2060, x2061, x2059, x2062, x2063, x2058, false)_>=_IF_4(x2060, x2061, x2059, x2062, x2063, x2058, empty(fstsplit(x2058, app(map_f(two, head(x2061)), x2059))))) & (\/x2068,x2069,x2070,x2071,x2072,x2073,x2074,x2075,x2076,x2077,x2078:length(x2066)=s(x2068) & leq(x2069, x2068)=false & (\/x2070,x2071,x2072,x2073,x2074:leq(x2069, x2068)=false & length(x2070)=x2068 ==> IF_2(x2071, x2072, x2070, x2073, x2074, x2069, false)_>=_IF_4(x2071, x2072, x2070, x2073, x2074, x2069, empty(fstsplit(x2069, app(map_f(two, head(x2072)), x2070))))) ==> IF_2(x2075, x2076, x2066, x2077, x2078, s(x2069), false)_>=_IF_4(x2075, x2076, x2066, x2077, x2078, s(x2069), empty(fstsplit(s(x2069), app(map_f(two, head(x2076)), x2066))))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) We simplified constraint (9) using rule (VI) where we applied the induction hypothesis (\/x2059,x2060,x2061,x2062,x2063:leq(x2058, x2057)=false & length(x2059)=x2057 ==> IF_2(x2060, x2061, x2059, x2062, x2063, x2058, false)_>=_IF_4(x2060, x2061, x2059, x2062, x2063, x2058, empty(fstsplit(x2058, app(map_f(two, head(x2061)), x2059))))) with sigma = [x2059 / x2066, x2060 / x1368, x2061 / x1369, x2062 / x1371, x2063 / x1372] which results in the following new constraint: (10) (IF_2(x1368, x1369, x2066, x1371, x1372, x2058, false)_>=_IF_4(x1368, x1369, x2066, x1371, x1372, x2058, empty(fstsplit(x2058, app(map_f(two, head(x1369)), x2066)))) & (\/x2068,x2069,x2070,x2071,x2072,x2073,x2074,x2075,x2076,x2077,x2078:length(x2066)=s(x2068) & leq(x2069, x2068)=false & (\/x2070,x2071,x2072,x2073,x2074:leq(x2069, x2068)=false & length(x2070)=x2068 ==> IF_2(x2071, x2072, x2070, x2073, x2074, x2069, false)_>=_IF_4(x2071, x2072, x2070, x2073, x2074, x2069, empty(fstsplit(x2069, app(map_f(two, head(x2072)), x2070))))) ==> IF_2(x2075, x2076, x2066, x2077, x2078, s(x2069), false)_>=_IF_4(x2075, x2076, x2066, x2077, x2078, s(x2069), empty(fstsplit(s(x2069), app(map_f(two, head(x2076)), x2066))))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) We simplified constraint (10) using rule (IV) which results in the following new constraint: (11) (IF_2(x1368, x1369, x2066, x1371, x1372, x2058, false)_>=_IF_4(x1368, x1369, x2066, x1371, x1372, x2058, empty(fstsplit(x2058, app(map_f(two, head(x1369)), x2066)))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) For Pair IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) the following chains were created: *We consider the chain IF_2(x1532, x1533, x1534, x1535, x1536, x1537, false) -> IF_4(x1532, x1533, x1534, x1535, x1536, x1537, empty(fstsplit(x1537, app(map_f(two, head(x1533)), x1534)))), IF_4(x1538, x1539, x1540, x1541, x1542, x1543, false) -> RING(x1538, tail(x1539), sndsplit(x1543, app(map_f(two, head(x1539)), x1540)), cons(fstsplit(x1543, app(map_f(two, head(x1539)), x1540)), x1541), x1542, x1543) which results in the following constraint: (1) (IF_4(x1532, x1533, x1534, x1535, x1536, x1537, empty(fstsplit(x1537, app(map_f(two, head(x1533)), x1534))))=IF_4(x1538, x1539, x1540, x1541, x1542, x1543, false) ==> IF_4(x1538, x1539, x1540, x1541, x1542, x1543, false)_>=_RING(x1538, tail(x1539), sndsplit(x1543, app(map_f(two, head(x1539)), x1540)), cons(fstsplit(x1543, app(map_f(two, head(x1539)), x1540)), x1541), x1542, x1543)) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (fstsplit(x1537, x2080)=x2079 & empty(x2079)=false ==> IF_4(x1532, x1533, x1534, x1535, x1536, x1537, false)_>=_RING(x1532, tail(x1533), sndsplit(x1537, app(map_f(two, head(x1533)), x1534)), cons(fstsplit(x1537, app(map_f(two, head(x1533)), x1534)), x1535), x1536, x1537)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x2079)=false which results in the following new constraint: (3) (false=false & fstsplit(x1537, x2080)=cons(x2085, x2084) ==> IF_4(x1532, x1533, x1534, x1535, x1536, x1537, false)_>=_RING(x1532, tail(x1533), sndsplit(x1537, app(map_f(two, head(x1533)), x1534)), cons(fstsplit(x1537, app(map_f(two, head(x1533)), x1534)), x1535), x1536, x1537)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (fstsplit(x1537, x2080)=cons(x2085, x2084) ==> IF_4(x1532, x1533, x1534, x1535, x1536, x1537, false)_>=_RING(x1532, tail(x1533), sndsplit(x1537, app(map_f(two, head(x1533)), x1534)), cons(fstsplit(x1537, app(map_f(two, head(x1533)), x1534)), x1535), x1536, x1537)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on fstsplit(x1537, x2080)=cons(x2085, x2084) which results in the following new constraint: (5) (cons(x2089, fstsplit(x2090, x2088))=cons(x2085, x2084) & (\/x2091,x2092,x2093,x2094,x2095,x2096,x2097:fstsplit(x2090, x2088)=cons(x2091, x2092) ==> IF_4(x2093, x2094, x2095, x2096, x2097, x2090, false)_>=_RING(x2093, tail(x2094), sndsplit(x2090, app(map_f(two, head(x2094)), x2095)), cons(fstsplit(x2090, app(map_f(two, head(x2094)), x2095)), x2096), x2097, x2090)) ==> IF_4(x1532, x1533, x1534, x1535, x1536, s(x2090), false)_>=_RING(x1532, tail(x1533), sndsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), cons(fstsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), x1535), x1536, s(x2090))) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (6) (IF_4(x1532, x1533, x1534, x1535, x1536, s(x2090), false)_>=_RING(x1532, tail(x1533), sndsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), cons(fstsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), x1535), x1536, s(x2090))) For Pair RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) the following chains were created: *We consider the chain IF_3(x1742, x1743, x1744, x1745, x1746, x1747, false) -> RING(x1742, x1743, sndsplit(x1747, x1744), cons(fstsplit(x1747, x1744), x1745), x1746, x1747), RING(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754) -> IF_9(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754, empty(map_f(three, x1751))) which results in the following constraint: (1) (RING(x1742, x1743, sndsplit(x1747, x1744), cons(fstsplit(x1747, x1744), x1745), x1746, x1747)=RING(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754) ==> RING(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754)_>=_IF_9(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754, empty(map_f(three, x1751)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x1742, x1743, x1750, cons(x1751, x1745), x1746, x1747)_>=_IF_9(x1742, x1743, x1750, cons(x1751, x1745), x1746, x1747, empty(map_f(three, x1751)))) *We consider the chain IF_7(x1780, x1781, x1782, x1783, x1784, x1785, false) -> RING(x1780, x1781, x1782, x1783, sndsplit(x1785, x1784), x1785), RING(x1786, x1787, x1788, cons(x1789, x1790), x1791, x1792) -> IF_9(x1786, x1787, x1788, cons(x1789, x1790), x1791, x1792, empty(map_f(three, x1789))) which results in the following constraint: (1) (RING(x1780, x1781, x1782, x1783, sndsplit(x1785, x1784), x1785)=RING(x1786, x1787, x1788, cons(x1789, x1790), x1791, x1792) ==> RING(x1786, x1787, x1788, cons(x1789, x1790), x1791, x1792)_>=_IF_9(x1786, x1787, x1788, cons(x1789, x1790), x1791, x1792, empty(map_f(three, x1789)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x1780, x1781, x1782, cons(x1789, x1790), x1791, x1785)_>=_IF_9(x1780, x1781, x1782, cons(x1789, x1790), x1791, x1785, empty(map_f(three, x1789)))) *We consider the chain IF_9(x1793, x1794, x1795, x1796, x1797, x1798, true) -> RING(x1793, x1794, x1795, tail(x1796), x1797, x1798), RING(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805) -> IF_9(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805, empty(map_f(three, x1802))) which results in the following constraint: (1) (RING(x1793, x1794, x1795, tail(x1796), x1797, x1798)=RING(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805) ==> RING(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805)_>=_IF_9(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805, empty(map_f(three, x1802)))) We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint: (2) (tail(x1796)=cons(x1802, x1803) ==> RING(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798)_>=_IF_9(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798, empty(map_f(three, x1802)))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on tail(x1796)=cons(x1802, x1803) which results in the following new constraint: (3) (x2110=cons(x1802, x1803) ==> RING(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798)_>=_IF_9(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798, empty(map_f(three, x1802)))) We simplified constraint (3) using rule (III) which results in the following new constraint: (4) (RING(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798)_>=_IF_9(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798, empty(map_f(three, x1802)))) *We consider the chain IF_8(x1812, x1813, x1814, x1815, x1816, x1817, false) -> RING(x1812, x1813, x1814, tail(x1815), sndsplit(x1817, app(map_f(three, head(x1815)), x1816)), x1817), RING(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824) -> IF_9(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824, empty(map_f(three, x1821))) which results in the following constraint: (1) (RING(x1812, x1813, x1814, tail(x1815), sndsplit(x1817, app(map_f(three, head(x1815)), x1816)), x1817)=RING(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824) ==> RING(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824)_>=_IF_9(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824, empty(map_f(three, x1821)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (tail(x1815)=cons(x1821, x1822) & app(x2113, x1816)=x2112 & sndsplit(x1817, x2112)=x1823 ==> RING(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817)_>=_IF_9(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817, empty(map_f(three, x1821)))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on tail(x1815)=cons(x1821, x1822) which results in the following new constraint: (3) (x2116=cons(x1821, x1822) & app(x2113, x1816)=x2112 & sndsplit(x1817, x2112)=x1823 ==> RING(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817)_>=_IF_9(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817, empty(map_f(three, x1821)))) We simplified constraint (3) using rules (III), (IV) which results in the following new constraint: (4) (RING(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817)_>=_IF_9(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817, empty(map_f(three, x1821)))) *We consider the chain IF_4(x1831, x1832, x1833, x1834, x1835, x1836, false) -> RING(x1831, tail(x1832), sndsplit(x1836, app(map_f(two, head(x1832)), x1833)), cons(fstsplit(x1836, app(map_f(two, head(x1832)), x1833)), x1834), x1835, x1836), RING(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843) -> IF_9(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843, empty(map_f(three, x1840))) which results in the following constraint: (1) (RING(x1831, tail(x1832), sndsplit(x1836, app(map_f(two, head(x1832)), x1833)), cons(fstsplit(x1836, app(map_f(two, head(x1832)), x1833)), x1834), x1835, x1836)=RING(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843) ==> RING(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843)_>=_IF_9(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843, empty(map_f(three, x1840)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x1831, x1838, x1839, cons(x1840, x1834), x1835, x1836)_>=_IF_9(x1831, x1838, x1839, cons(x1840, x1834), x1835, x1836, empty(map_f(three, x1840)))) To summarize, we get the following constraints P__>=_ for the following pairs. *RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) *(RING(x274, x275, x282, x283, x278, x279)_>=_IF_2(x274, x275, x282, x283, x278, x279, leq(x279, length(x282)))) *(RING(x310, x311, x312, x313, x320, x315)_>=_IF_2(x310, x311, x312, x313, x320, x315, leq(x315, length(x312)))) *(RING(x322, x323, x324, x331, x326, x327)_>=_IF_2(x322, x323, x324, x331, x326, x327, leq(x327, length(x324)))) *(RING(x340, x341, x342, x349, x350, x345)_>=_IF_2(x340, x341, x342, x349, x350, x345, leq(x345, length(x342)))) *(RING(x358, x365, x366, x367, x362, x363)_>=_IF_2(x358, x365, x366, x367, x362, x363, leq(x363, length(x366)))) *IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) *(IF_2(x396, x397, x398, x399, x400, 0, true)_>=_IF_3(x396, x397, x398, x399, x400, 0, empty(fstsplit(0, x398)))) *(IF_2(x396, x397, x1898, x399, x400, x1892, true)_>=_IF_3(x396, x397, x1898, x399, x400, x1892, empty(fstsplit(x1892, x1898))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) *IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) *(IF_3(x506, x507, cons(x1917, x1916), x509, x510, s(x1918), false)_>=_RING(x506, x507, sndsplit(s(x1918), cons(x1917, x1916)), cons(fstsplit(s(x1918), cons(x1917, x1916)), x509), x510, s(x1918))) *RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) *(RING(x726, x727, x734, x735, x730, x731)_>=_IF_6(x726, x727, x734, x735, x730, x731, leq(x731, length(x730)))) *(RING(x762, x763, x764, x765, x772, x767)_>=_IF_6(x762, x763, x764, x765, x772, x767, leq(x767, length(x772)))) *(RING(x774, x775, x776, x783, x778, x779)_>=_IF_6(x774, x775, x776, x783, x778, x779, leq(x779, length(x778)))) *(RING(x792, x793, x794, x801, x802, x797)_>=_IF_6(x792, x793, x794, x801, x802, x797, leq(x797, length(x802)))) *(RING(x810, x817, x818, x819, x814, x815)_>=_IF_6(x810, x817, x818, x819, x814, x815, leq(x815, length(x814)))) *IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) *(IF_6(x872, x873, x874, x875, x876, 0, true)_>=_IF_7(x872, x873, x874, x875, x876, 0, empty(fstsplit(0, x876)))) *(IF_6(x872, x873, x874, x875, x1963, x1957, true)_>=_IF_7(x872, x873, x874, x875, x1963, x1957, empty(fstsplit(x1957, x1963))) ==> IF_6(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), true)_>=_IF_7(x872, x873, x874, x875, cons(x1964, x1963), s(x1957), empty(fstsplit(s(x1957), cons(x1964, x1963))))) *IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) *(IF_7(x982, x983, x984, x985, cons(x1982, x1981), s(x1983), false)_>=_RING(x982, x983, x984, x985, sndsplit(s(x1983), cons(x1982, x1981)), s(x1983))) *IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) *(IF_9(x1135, x1136, x1137, cons(nil, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(nil, x1139)), x1140, x1141)) *IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) *(IF_6(x1184, x1185, x1186, x1187, nil, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, nil, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), nil))))) *(IF_6(x1184, x1185, x1186, x1187, x2022, x2014, false)_>=_IF_8(x1184, x1185, x1186, x1187, x2022, x2014, empty(fstsplit(x2014, app(map_f(three, head(x1187)), x2022)))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) *IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) *(IF_8(x1312, x1313, x1314, x1315, x1316, s(x2046), false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(s(x2046), app(map_f(three, head(x1315)), x1316)), s(x2046))) *IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) *(IF_2(x1368, x1369, nil, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, nil, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), nil))))) *(IF_2(x1368, x1369, x2066, x1371, x1372, x2058, false)_>=_IF_4(x1368, x1369, x2066, x1371, x1372, x2058, empty(fstsplit(x2058, app(map_f(two, head(x1369)), x2066)))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) *IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) *(IF_4(x1532, x1533, x1534, x1535, x1536, s(x2090), false)_>=_RING(x1532, tail(x1533), sndsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), cons(fstsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), x1535), x1536, s(x2090))) *RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) *(RING(x1742, x1743, x1750, cons(x1751, x1745), x1746, x1747)_>=_IF_9(x1742, x1743, x1750, cons(x1751, x1745), x1746, x1747, empty(map_f(three, x1751)))) *(RING(x1780, x1781, x1782, cons(x1789, x1790), x1791, x1785)_>=_IF_9(x1780, x1781, x1782, cons(x1789, x1790), x1791, x1785, empty(map_f(three, x1789)))) *(RING(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798)_>=_IF_9(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798, empty(map_f(three, x1802)))) *(RING(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817)_>=_IF_9(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817, empty(map_f(three, x1821)))) *(RING(x1831, x1838, x1839, cons(x1840, x1834), x1835, x1836)_>=_IF_9(x1831, x1838, x1839, cons(x1840, x1834), x1835, x1836, empty(map_f(three, x1840)))) The constraints for P_> respective P_bound are constructed from P__>=_ where we just replace every occurence of "t _>=_ s" in P__>=_ by "t > s" respective "t _>=_ c". Here c stands for the fresh constant used for P_bound. Using the following integer polynomial ordering the resulting constraints can be solved Polynomial interpretation [NONINF]: POL(0) = 0 POL(IF_2(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 - x_5 - x_6 - x_7 POL(IF_3(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 - x_5 - x_6 POL(IF_4(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 - x_5 - x_6 POL(IF_6(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 - x_5 - x_6 - x_7 POL(IF_7(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 - x_5 - x_6 + x_7 POL(IF_8(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = -1 + x_1 - x_6 POL(IF_9(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 - x_4 - x_5 - x_6 + x_7 POL(RING(x_1, x_2, x_3, x_4, x_5, x_6)) = x_1 - x_5 - x_6 POL(app(x_1, x_2)) = 0 POL(c) = -2 POL(cons(x_1, x_2)) = 0 POL(empty(x_1)) = 0 POL(f(x_1, x_2)) = x_1 + x_2 POL(false) = 0 POL(fstsplit(x_1, x_2)) = 0 POL(head(x_1)) = 0 POL(length(x_1)) = 0 POL(leq(x_1, x_2)) = 0 POL(map_f(x_1, x_2)) = 0 POL(nil) = 1 POL(s(x_1)) = 1 POL(sndsplit(x_1, x_2)) = x_1 POL(tail(x_1)) = 0 POL(three) = 0 POL(true) = 0 POL(two) = 0 The following pairs are in P_>: IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) The following pairs are in P_bound: IF_7(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) The following rules are usable: true -> leq(0, m) false -> leq(s(n), 0) leq(n, m) -> leq(s(n), s(m)) empty(nil) -> true empty(cons(h, t)) -> false x -> sndsplit(0, x) nil -> sndsplit(s(n), nil) ---------------------------------------- (67) Obligation: Q DP problem: The TRS P consists of the following rules: RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3))) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (68) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (69) Obligation: Q DP problem: The TRS P consists of the following rules: IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (70) NonInfProof (EQUIVALENT) The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: Note that final constraints are written in bold face. For Pair RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) the following chains were created: *We consider the chain IF_3(x274, x275, x276, x277, x278, x279, false) -> RING(x274, x275, sndsplit(x279, x276), cons(fstsplit(x279, x276), x277), x278, x279), RING(x280, x281, x282, x283, x284, x285) -> IF_2(x280, x281, x282, x283, x284, x285, leq(x285, length(x282))) which results in the following constraint: (1) (RING(x274, x275, sndsplit(x279, x276), cons(fstsplit(x279, x276), x277), x278, x279)=RING(x280, x281, x282, x283, x284, x285) ==> RING(x280, x281, x282, x283, x284, x285)_>=_IF_2(x280, x281, x282, x283, x284, x285, leq(x285, length(x282)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x274, x275, x282, x283, x278, x279)_>=_IF_2(x274, x275, x282, x283, x278, x279, leq(x279, length(x282)))) *We consider the chain IF_9(x322, x323, x324, x325, x326, x327, true) -> RING(x322, x323, x324, tail(x325), x326, x327), RING(x328, x329, x330, x331, x332, x333) -> IF_2(x328, x329, x330, x331, x332, x333, leq(x333, length(x330))) which results in the following constraint: (1) (RING(x322, x323, x324, tail(x325), x326, x327)=RING(x328, x329, x330, x331, x332, x333) ==> RING(x328, x329, x330, x331, x332, x333)_>=_IF_2(x328, x329, x330, x331, x332, x333, leq(x333, length(x330)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x322, x323, x324, x331, x326, x327)_>=_IF_2(x322, x323, x324, x331, x326, x327, leq(x327, length(x324)))) *We consider the chain IF_8(x340, x341, x342, x343, x344, x345, false) -> RING(x340, x341, x342, tail(x343), sndsplit(x345, app(map_f(three, head(x343)), x344)), x345), RING(x346, x347, x348, x349, x350, x351) -> IF_2(x346, x347, x348, x349, x350, x351, leq(x351, length(x348))) which results in the following constraint: (1) (RING(x340, x341, x342, tail(x343), sndsplit(x345, app(map_f(three, head(x343)), x344)), x345)=RING(x346, x347, x348, x349, x350, x351) ==> RING(x346, x347, x348, x349, x350, x351)_>=_IF_2(x346, x347, x348, x349, x350, x351, leq(x351, length(x348)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x340, x341, x342, x349, x350, x345)_>=_IF_2(x340, x341, x342, x349, x350, x345, leq(x345, length(x342)))) *We consider the chain IF_4(x358, x359, x360, x361, x362, x363, false) -> RING(x358, tail(x359), sndsplit(x363, app(map_f(two, head(x359)), x360)), cons(fstsplit(x363, app(map_f(two, head(x359)), x360)), x361), x362, x363), RING(x364, x365, x366, x367, x368, x369) -> IF_2(x364, x365, x366, x367, x368, x369, leq(x369, length(x366))) which results in the following constraint: (1) (RING(x358, tail(x359), sndsplit(x363, app(map_f(two, head(x359)), x360)), cons(fstsplit(x363, app(map_f(two, head(x359)), x360)), x361), x362, x363)=RING(x364, x365, x366, x367, x368, x369) ==> RING(x364, x365, x366, x367, x368, x369)_>=_IF_2(x364, x365, x366, x367, x368, x369, leq(x369, length(x366)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x358, x365, x366, x367, x362, x363)_>=_IF_2(x358, x365, x366, x367, x362, x363, leq(x363, length(x366)))) For Pair IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) the following chains were created: *We consider the chain RING(x396, x397, x398, x399, x400, x401) -> IF_2(x396, x397, x398, x399, x400, x401, leq(x401, length(x398))), IF_2(x402, x403, x404, x405, x406, x407, true) -> IF_3(x402, x403, x404, x405, x406, x407, empty(fstsplit(x407, x404))) which results in the following constraint: (1) (IF_2(x396, x397, x398, x399, x400, x401, leq(x401, length(x398)))=IF_2(x402, x403, x404, x405, x406, x407, true) ==> IF_2(x402, x403, x404, x405, x406, x407, true)_>=_IF_3(x402, x403, x404, x405, x406, x407, empty(fstsplit(x407, x404)))) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (length(x398)=x1888 & leq(x401, x1888)=true ==> IF_2(x396, x397, x398, x399, x400, x401, true)_>=_IF_3(x396, x397, x398, x399, x400, x401, empty(fstsplit(x401, x398)))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on leq(x401, x1888)=true which results in the following new constraints: (3) (true=true & length(x398)=x1889 ==> IF_2(x396, x397, x398, x399, x400, 0, true)_>=_IF_3(x396, x397, x398, x399, x400, 0, empty(fstsplit(0, x398)))) (4) (leq(x1892, x1891)=true & length(x398)=s(x1891) & (\/x1893,x1894,x1895,x1896,x1897:leq(x1892, x1891)=true & length(x1893)=x1891 ==> IF_2(x1894, x1895, x1893, x1896, x1897, x1892, true)_>=_IF_3(x1894, x1895, x1893, x1896, x1897, x1892, empty(fstsplit(x1892, x1893)))) ==> IF_2(x396, x397, x398, x399, x400, s(x1892), true)_>=_IF_3(x396, x397, x398, x399, x400, s(x1892), empty(fstsplit(s(x1892), x398)))) We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint: (5) (IF_2(x396, x397, x398, x399, x400, 0, true)_>=_IF_3(x396, x397, x398, x399, x400, 0, empty(fstsplit(0, x398)))) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on length(x398)=s(x1891) which results in the following new constraint: (6) (s(length(x1898))=s(x1891) & leq(x1892, x1891)=true & (\/x1893,x1894,x1895,x1896,x1897:leq(x1892, x1891)=true & length(x1893)=x1891 ==> IF_2(x1894, x1895, x1893, x1896, x1897, x1892, true)_>=_IF_3(x1894, x1895, x1893, x1896, x1897, x1892, empty(fstsplit(x1892, x1893)))) & (\/x1900,x1901,x1902,x1903,x1904,x1905,x1906,x1907,x1908,x1909,x1910:length(x1898)=s(x1900) & leq(x1901, x1900)=true & (\/x1902,x1903,x1904,x1905,x1906:leq(x1901, x1900)=true & length(x1902)=x1900 ==> IF_2(x1903, x1904, x1902, x1905, x1906, x1901, true)_>=_IF_3(x1903, x1904, x1902, x1905, x1906, x1901, empty(fstsplit(x1901, x1902)))) ==> IF_2(x1907, x1908, x1898, x1909, x1910, s(x1901), true)_>=_IF_3(x1907, x1908, x1898, x1909, x1910, s(x1901), empty(fstsplit(s(x1901), x1898)))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) We simplified constraint (6) using rules (I), (II) which results in the following new constraint: (7) (length(x1898)=x1891 & leq(x1892, x1891)=true & (\/x1893,x1894,x1895,x1896,x1897:leq(x1892, x1891)=true & length(x1893)=x1891 ==> IF_2(x1894, x1895, x1893, x1896, x1897, x1892, true)_>=_IF_3(x1894, x1895, x1893, x1896, x1897, x1892, empty(fstsplit(x1892, x1893)))) & (\/x1900,x1901,x1902,x1903,x1904,x1905,x1906,x1907,x1908,x1909,x1910:length(x1898)=s(x1900) & leq(x1901, x1900)=true & (\/x1902,x1903,x1904,x1905,x1906:leq(x1901, x1900)=true & length(x1902)=x1900 ==> IF_2(x1903, x1904, x1902, x1905, x1906, x1901, true)_>=_IF_3(x1903, x1904, x1902, x1905, x1906, x1901, empty(fstsplit(x1901, x1902)))) ==> IF_2(x1907, x1908, x1898, x1909, x1910, s(x1901), true)_>=_IF_3(x1907, x1908, x1898, x1909, x1910, s(x1901), empty(fstsplit(s(x1901), x1898)))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) We simplified constraint (7) using rule (VI) where we applied the induction hypothesis (\/x1893,x1894,x1895,x1896,x1897:leq(x1892, x1891)=true & length(x1893)=x1891 ==> IF_2(x1894, x1895, x1893, x1896, x1897, x1892, true)_>=_IF_3(x1894, x1895, x1893, x1896, x1897, x1892, empty(fstsplit(x1892, x1893)))) with sigma = [x1893 / x1898, x1894 / x396, x1895 / x397, x1896 / x399, x1897 / x400] which results in the following new constraint: (8) (IF_2(x396, x397, x1898, x399, x400, x1892, true)_>=_IF_3(x396, x397, x1898, x399, x400, x1892, empty(fstsplit(x1892, x1898))) & (\/x1900,x1901,x1902,x1903,x1904,x1905,x1906,x1907,x1908,x1909,x1910:length(x1898)=s(x1900) & leq(x1901, x1900)=true & (\/x1902,x1903,x1904,x1905,x1906:leq(x1901, x1900)=true & length(x1902)=x1900 ==> IF_2(x1903, x1904, x1902, x1905, x1906, x1901, true)_>=_IF_3(x1903, x1904, x1902, x1905, x1906, x1901, empty(fstsplit(x1901, x1902)))) ==> IF_2(x1907, x1908, x1898, x1909, x1910, s(x1901), true)_>=_IF_3(x1907, x1908, x1898, x1909, x1910, s(x1901), empty(fstsplit(s(x1901), x1898)))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) We simplified constraint (8) using rule (IV) which results in the following new constraint: (9) (IF_2(x396, x397, x1898, x399, x400, x1892, true)_>=_IF_3(x396, x397, x1898, x399, x400, x1892, empty(fstsplit(x1892, x1898))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) For Pair IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) the following chains were created: *We consider the chain IF_2(x506, x507, x508, x509, x510, x511, true) -> IF_3(x506, x507, x508, x509, x510, x511, empty(fstsplit(x511, x508))), IF_3(x512, x513, x514, x515, x516, x517, false) -> RING(x512, x513, sndsplit(x517, x514), cons(fstsplit(x517, x514), x515), x516, x517) which results in the following constraint: (1) (IF_3(x506, x507, x508, x509, x510, x511, empty(fstsplit(x511, x508)))=IF_3(x512, x513, x514, x515, x516, x517, false) ==> IF_3(x512, x513, x514, x515, x516, x517, false)_>=_RING(x512, x513, sndsplit(x517, x514), cons(fstsplit(x517, x514), x515), x516, x517)) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (fstsplit(x511, x508)=x1911 & empty(x1911)=false ==> IF_3(x506, x507, x508, x509, x510, x511, false)_>=_RING(x506, x507, sndsplit(x511, x508), cons(fstsplit(x511, x508), x509), x510, x511)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x1911)=false which results in the following new constraint: (3) (false=false & fstsplit(x511, x508)=cons(x1913, x1912) ==> IF_3(x506, x507, x508, x509, x510, x511, false)_>=_RING(x506, x507, sndsplit(x511, x508), cons(fstsplit(x511, x508), x509), x510, x511)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (fstsplit(x511, x508)=cons(x1913, x1912) ==> IF_3(x506, x507, x508, x509, x510, x511, false)_>=_RING(x506, x507, sndsplit(x511, x508), cons(fstsplit(x511, x508), x509), x510, x511)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on fstsplit(x511, x508)=cons(x1913, x1912) which results in the following new constraint: (5) (cons(x1917, fstsplit(x1918, x1916))=cons(x1913, x1912) & (\/x1919,x1920,x1921,x1922,x1923,x1924:fstsplit(x1918, x1916)=cons(x1919, x1920) ==> IF_3(x1921, x1922, x1916, x1923, x1924, x1918, false)_>=_RING(x1921, x1922, sndsplit(x1918, x1916), cons(fstsplit(x1918, x1916), x1923), x1924, x1918)) ==> IF_3(x506, x507, cons(x1917, x1916), x509, x510, s(x1918), false)_>=_RING(x506, x507, sndsplit(s(x1918), cons(x1917, x1916)), cons(fstsplit(s(x1918), cons(x1917, x1916)), x509), x510, s(x1918))) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (6) (IF_3(x506, x507, cons(x1917, x1916), x509, x510, s(x1918), false)_>=_RING(x506, x507, sndsplit(s(x1918), cons(x1917, x1916)), cons(fstsplit(s(x1918), cons(x1917, x1916)), x509), x510, s(x1918))) For Pair RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) the following chains were created: *We consider the chain IF_3(x726, x727, x728, x729, x730, x731, false) -> RING(x726, x727, sndsplit(x731, x728), cons(fstsplit(x731, x728), x729), x730, x731), RING(x732, x733, x734, x735, x736, x737) -> IF_6(x732, x733, x734, x735, x736, x737, leq(x737, length(x736))) which results in the following constraint: (1) (RING(x726, x727, sndsplit(x731, x728), cons(fstsplit(x731, x728), x729), x730, x731)=RING(x732, x733, x734, x735, x736, x737) ==> RING(x732, x733, x734, x735, x736, x737)_>=_IF_6(x732, x733, x734, x735, x736, x737, leq(x737, length(x736)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x726, x727, x734, x735, x730, x731)_>=_IF_6(x726, x727, x734, x735, x730, x731, leq(x731, length(x730)))) *We consider the chain IF_9(x774, x775, x776, x777, x778, x779, true) -> RING(x774, x775, x776, tail(x777), x778, x779), RING(x780, x781, x782, x783, x784, x785) -> IF_6(x780, x781, x782, x783, x784, x785, leq(x785, length(x784))) which results in the following constraint: (1) (RING(x774, x775, x776, tail(x777), x778, x779)=RING(x780, x781, x782, x783, x784, x785) ==> RING(x780, x781, x782, x783, x784, x785)_>=_IF_6(x780, x781, x782, x783, x784, x785, leq(x785, length(x784)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x774, x775, x776, x783, x778, x779)_>=_IF_6(x774, x775, x776, x783, x778, x779, leq(x779, length(x778)))) *We consider the chain IF_8(x792, x793, x794, x795, x796, x797, false) -> RING(x792, x793, x794, tail(x795), sndsplit(x797, app(map_f(three, head(x795)), x796)), x797), RING(x798, x799, x800, x801, x802, x803) -> IF_6(x798, x799, x800, x801, x802, x803, leq(x803, length(x802))) which results in the following constraint: (1) (RING(x792, x793, x794, tail(x795), sndsplit(x797, app(map_f(three, head(x795)), x796)), x797)=RING(x798, x799, x800, x801, x802, x803) ==> RING(x798, x799, x800, x801, x802, x803)_>=_IF_6(x798, x799, x800, x801, x802, x803, leq(x803, length(x802)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x792, x793, x794, x801, x802, x797)_>=_IF_6(x792, x793, x794, x801, x802, x797, leq(x797, length(x802)))) *We consider the chain IF_4(x810, x811, x812, x813, x814, x815, false) -> RING(x810, tail(x811), sndsplit(x815, app(map_f(two, head(x811)), x812)), cons(fstsplit(x815, app(map_f(two, head(x811)), x812)), x813), x814, x815), RING(x816, x817, x818, x819, x820, x821) -> IF_6(x816, x817, x818, x819, x820, x821, leq(x821, length(x820))) which results in the following constraint: (1) (RING(x810, tail(x811), sndsplit(x815, app(map_f(two, head(x811)), x812)), cons(fstsplit(x815, app(map_f(two, head(x811)), x812)), x813), x814, x815)=RING(x816, x817, x818, x819, x820, x821) ==> RING(x816, x817, x818, x819, x820, x821)_>=_IF_6(x816, x817, x818, x819, x820, x821, leq(x821, length(x820)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x810, x817, x818, x819, x814, x815)_>=_IF_6(x810, x817, x818, x819, x814, x815, leq(x815, length(x814)))) For Pair IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) the following chains were created: *We consider the chain RING(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141) -> IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, empty(map_f(three, x1138))), IF_9(x1142, x1143, x1144, x1145, x1146, x1147, true) -> RING(x1142, x1143, x1144, tail(x1145), x1146, x1147) which results in the following constraint: (1) (IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, empty(map_f(three, x1138)))=IF_9(x1142, x1143, x1144, x1145, x1146, x1147, true) ==> IF_9(x1142, x1143, x1144, x1145, x1146, x1147, true)_>=_RING(x1142, x1143, x1144, tail(x1145), x1146, x1147)) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (three=x1991 & map_f(x1991, x1138)=x1990 & empty(x1990)=true ==> IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(x1138, x1139)), x1140, x1141)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x1990)=true which results in the following new constraint: (3) (true=true & three=x1991 & map_f(x1991, x1138)=nil ==> IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(x1138, x1139)), x1140, x1141)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (three=x1991 & map_f(x1991, x1138)=nil ==> IF_9(x1135, x1136, x1137, cons(x1138, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(x1138, x1139)), x1140, x1141)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on map_f(x1991, x1138)=nil which results in the following new constraints: (5) (nil=nil & three=x1994 ==> IF_9(x1135, x1136, x1137, cons(nil, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(nil, x1139)), x1140, x1141)) (6) (app(f(x1997, x1996), map_f(x1997, x1995))=nil & three=x1997 & (\/x1998,x1999,x2000,x2001,x2002,x2003:map_f(x1997, x1995)=nil & three=x1997 ==> IF_9(x1998, x1999, x2000, cons(x1995, x2001), x2002, x2003, true)_>=_RING(x1998, x1999, x2000, tail(cons(x1995, x2001)), x2002, x2003)) ==> IF_9(x1135, x1136, x1137, cons(cons(x1996, x1995), x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(cons(x1996, x1995), x1139)), x1140, x1141)) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (7) (IF_9(x1135, x1136, x1137, cons(nil, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(nil, x1139)), x1140, x1141)) We simplified constraint (6) using rules (IV), (VII) which results in the following new constraint: (8) (f(x1997, x1996)=x2004 & map_f(x1997, x1995)=x2005 & app(x2004, x2005)=nil & three=x1997 ==> IF_9(x1135, x1136, x1137, cons(cons(x1996, x1995), x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(cons(x1996, x1995), x1139)), x1140, x1141)) We simplified constraint (8) using rule (V) (with possible (I) afterwards) using induction on app(x2004, x2005)=nil which results in the following new constraint: (9) (x2006=nil & f(x1997, x1996)=nil & map_f(x1997, x1995)=x2006 & three=x1997 ==> IF_9(x1135, x1136, x1137, cons(cons(x1996, x1995), x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(cons(x1996, x1995), x1139)), x1140, x1141)) We solved constraint (9) using rules (I), (II). For Pair IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) the following chains were created: *We consider the chain RING(x1184, x1185, x1186, x1187, x1188, x1189) -> IF_6(x1184, x1185, x1186, x1187, x1188, x1189, leq(x1189, length(x1188))), IF_6(x1190, x1191, x1192, x1193, x1194, x1195, false) -> IF_8(x1190, x1191, x1192, x1193, x1194, x1195, empty(fstsplit(x1195, app(map_f(three, head(x1193)), x1194)))) which results in the following constraint: (1) (IF_6(x1184, x1185, x1186, x1187, x1188, x1189, leq(x1189, length(x1188)))=IF_6(x1190, x1191, x1192, x1193, x1194, x1195, false) ==> IF_6(x1190, x1191, x1192, x1193, x1194, x1195, false)_>=_IF_8(x1190, x1191, x1192, x1193, x1194, x1195, empty(fstsplit(x1195, app(map_f(three, head(x1193)), x1194))))) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (length(x1188)=x2010 & leq(x1189, x2010)=false ==> IF_6(x1184, x1185, x1186, x1187, x1188, x1189, false)_>=_IF_8(x1184, x1185, x1186, x1187, x1188, x1189, empty(fstsplit(x1189, app(map_f(three, head(x1187)), x1188))))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on leq(x1189, x2010)=false which results in the following new constraints: (3) (false=false & length(x1188)=0 ==> IF_6(x1184, x1185, x1186, x1187, x1188, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, x1188, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), x1188))))) (4) (leq(x2014, x2013)=false & length(x1188)=s(x2013) & (\/x2015,x2016,x2017,x2018,x2019:leq(x2014, x2013)=false & length(x2015)=x2013 ==> IF_6(x2016, x2017, x2018, x2019, x2015, x2014, false)_>=_IF_8(x2016, x2017, x2018, x2019, x2015, x2014, empty(fstsplit(x2014, app(map_f(three, head(x2019)), x2015))))) ==> IF_6(x1184, x1185, x1186, x1187, x1188, s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, x1188, s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), x1188))))) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (5) (length(x1188)=0 ==> IF_6(x1184, x1185, x1186, x1187, x1188, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, x1188, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), x1188))))) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on length(x1188)=s(x2013) which results in the following new constraint: (6) (s(length(x2022))=s(x2013) & leq(x2014, x2013)=false & (\/x2015,x2016,x2017,x2018,x2019:leq(x2014, x2013)=false & length(x2015)=x2013 ==> IF_6(x2016, x2017, x2018, x2019, x2015, x2014, false)_>=_IF_8(x2016, x2017, x2018, x2019, x2015, x2014, empty(fstsplit(x2014, app(map_f(three, head(x2019)), x2015))))) & (\/x2024,x2025,x2026,x2027,x2028,x2029,x2030,x2031,x2032,x2033,x2034:length(x2022)=s(x2024) & leq(x2025, x2024)=false & (\/x2026,x2027,x2028,x2029,x2030:leq(x2025, x2024)=false & length(x2026)=x2024 ==> IF_6(x2027, x2028, x2029, x2030, x2026, x2025, false)_>=_IF_8(x2027, x2028, x2029, x2030, x2026, x2025, empty(fstsplit(x2025, app(map_f(three, head(x2030)), x2026))))) ==> IF_6(x2031, x2032, x2033, x2034, x2022, s(x2025), false)_>=_IF_8(x2031, x2032, x2033, x2034, x2022, s(x2025), empty(fstsplit(s(x2025), app(map_f(three, head(x2034)), x2022))))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on length(x1188)=0 which results in the following new constraint: (7) (0=0 ==> IF_6(x1184, x1185, x1186, x1187, nil, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, nil, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), nil))))) We simplified constraint (7) using rules (I), (II) which results in the following new constraint: (8) (IF_6(x1184, x1185, x1186, x1187, nil, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, nil, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), nil))))) We simplified constraint (6) using rules (I), (II) which results in the following new constraint: (9) (length(x2022)=x2013 & leq(x2014, x2013)=false & (\/x2015,x2016,x2017,x2018,x2019:leq(x2014, x2013)=false & length(x2015)=x2013 ==> IF_6(x2016, x2017, x2018, x2019, x2015, x2014, false)_>=_IF_8(x2016, x2017, x2018, x2019, x2015, x2014, empty(fstsplit(x2014, app(map_f(three, head(x2019)), x2015))))) & (\/x2024,x2025,x2026,x2027,x2028,x2029,x2030,x2031,x2032,x2033,x2034:length(x2022)=s(x2024) & leq(x2025, x2024)=false & (\/x2026,x2027,x2028,x2029,x2030:leq(x2025, x2024)=false & length(x2026)=x2024 ==> IF_6(x2027, x2028, x2029, x2030, x2026, x2025, false)_>=_IF_8(x2027, x2028, x2029, x2030, x2026, x2025, empty(fstsplit(x2025, app(map_f(three, head(x2030)), x2026))))) ==> IF_6(x2031, x2032, x2033, x2034, x2022, s(x2025), false)_>=_IF_8(x2031, x2032, x2033, x2034, x2022, s(x2025), empty(fstsplit(s(x2025), app(map_f(three, head(x2034)), x2022))))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) We simplified constraint (9) using rule (VI) where we applied the induction hypothesis (\/x2015,x2016,x2017,x2018,x2019:leq(x2014, x2013)=false & length(x2015)=x2013 ==> IF_6(x2016, x2017, x2018, x2019, x2015, x2014, false)_>=_IF_8(x2016, x2017, x2018, x2019, x2015, x2014, empty(fstsplit(x2014, app(map_f(three, head(x2019)), x2015))))) with sigma = [x2015 / x2022, x2016 / x1184, x2017 / x1185, x2018 / x1186, x2019 / x1187] which results in the following new constraint: (10) (IF_6(x1184, x1185, x1186, x1187, x2022, x2014, false)_>=_IF_8(x1184, x1185, x1186, x1187, x2022, x2014, empty(fstsplit(x2014, app(map_f(three, head(x1187)), x2022)))) & (\/x2024,x2025,x2026,x2027,x2028,x2029,x2030,x2031,x2032,x2033,x2034:length(x2022)=s(x2024) & leq(x2025, x2024)=false & (\/x2026,x2027,x2028,x2029,x2030:leq(x2025, x2024)=false & length(x2026)=x2024 ==> IF_6(x2027, x2028, x2029, x2030, x2026, x2025, false)_>=_IF_8(x2027, x2028, x2029, x2030, x2026, x2025, empty(fstsplit(x2025, app(map_f(three, head(x2030)), x2026))))) ==> IF_6(x2031, x2032, x2033, x2034, x2022, s(x2025), false)_>=_IF_8(x2031, x2032, x2033, x2034, x2022, s(x2025), empty(fstsplit(s(x2025), app(map_f(three, head(x2034)), x2022))))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) We simplified constraint (10) using rule (IV) which results in the following new constraint: (11) (IF_6(x1184, x1185, x1186, x1187, x2022, x2014, false)_>=_IF_8(x1184, x1185, x1186, x1187, x2022, x2014, empty(fstsplit(x2014, app(map_f(three, head(x1187)), x2022)))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) For Pair IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) the following chains were created: *We consider the chain IF_6(x1312, x1313, x1314, x1315, x1316, x1317, false) -> IF_8(x1312, x1313, x1314, x1315, x1316, x1317, empty(fstsplit(x1317, app(map_f(three, head(x1315)), x1316)))), IF_8(x1318, x1319, x1320, x1321, x1322, x1323, false) -> RING(x1318, x1319, x1320, tail(x1321), sndsplit(x1323, app(map_f(three, head(x1321)), x1322)), x1323) which results in the following constraint: (1) (IF_8(x1312, x1313, x1314, x1315, x1316, x1317, empty(fstsplit(x1317, app(map_f(three, head(x1315)), x1316))))=IF_8(x1318, x1319, x1320, x1321, x1322, x1323, false) ==> IF_8(x1318, x1319, x1320, x1321, x1322, x1323, false)_>=_RING(x1318, x1319, x1320, tail(x1321), sndsplit(x1323, app(map_f(three, head(x1321)), x1322)), x1323)) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (fstsplit(x1317, x2036)=x2035 & empty(x2035)=false ==> IF_8(x1312, x1313, x1314, x1315, x1316, x1317, false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(x1317, app(map_f(three, head(x1315)), x1316)), x1317)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x2035)=false which results in the following new constraint: (3) (false=false & fstsplit(x1317, x2036)=cons(x2041, x2040) ==> IF_8(x1312, x1313, x1314, x1315, x1316, x1317, false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(x1317, app(map_f(three, head(x1315)), x1316)), x1317)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (fstsplit(x1317, x2036)=cons(x2041, x2040) ==> IF_8(x1312, x1313, x1314, x1315, x1316, x1317, false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(x1317, app(map_f(three, head(x1315)), x1316)), x1317)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on fstsplit(x1317, x2036)=cons(x2041, x2040) which results in the following new constraint: (5) (cons(x2045, fstsplit(x2046, x2044))=cons(x2041, x2040) & (\/x2047,x2048,x2049,x2050,x2051,x2052,x2053:fstsplit(x2046, x2044)=cons(x2047, x2048) ==> IF_8(x2049, x2050, x2051, x2052, x2053, x2046, false)_>=_RING(x2049, x2050, x2051, tail(x2052), sndsplit(x2046, app(map_f(three, head(x2052)), x2053)), x2046)) ==> IF_8(x1312, x1313, x1314, x1315, x1316, s(x2046), false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(s(x2046), app(map_f(three, head(x1315)), x1316)), s(x2046))) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (6) (IF_8(x1312, x1313, x1314, x1315, x1316, s(x2046), false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(s(x2046), app(map_f(three, head(x1315)), x1316)), s(x2046))) For Pair IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) the following chains were created: *We consider the chain RING(x1368, x1369, x1370, x1371, x1372, x1373) -> IF_2(x1368, x1369, x1370, x1371, x1372, x1373, leq(x1373, length(x1370))), IF_2(x1374, x1375, x1376, x1377, x1378, x1379, false) -> IF_4(x1374, x1375, x1376, x1377, x1378, x1379, empty(fstsplit(x1379, app(map_f(two, head(x1375)), x1376)))) which results in the following constraint: (1) (IF_2(x1368, x1369, x1370, x1371, x1372, x1373, leq(x1373, length(x1370)))=IF_2(x1374, x1375, x1376, x1377, x1378, x1379, false) ==> IF_2(x1374, x1375, x1376, x1377, x1378, x1379, false)_>=_IF_4(x1374, x1375, x1376, x1377, x1378, x1379, empty(fstsplit(x1379, app(map_f(two, head(x1375)), x1376))))) We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint: (2) (length(x1370)=x2054 & leq(x1373, x2054)=false ==> IF_2(x1368, x1369, x1370, x1371, x1372, x1373, false)_>=_IF_4(x1368, x1369, x1370, x1371, x1372, x1373, empty(fstsplit(x1373, app(map_f(two, head(x1369)), x1370))))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on leq(x1373, x2054)=false which results in the following new constraints: (3) (false=false & length(x1370)=0 ==> IF_2(x1368, x1369, x1370, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, x1370, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), x1370))))) (4) (leq(x2058, x2057)=false & length(x1370)=s(x2057) & (\/x2059,x2060,x2061,x2062,x2063:leq(x2058, x2057)=false & length(x2059)=x2057 ==> IF_2(x2060, x2061, x2059, x2062, x2063, x2058, false)_>=_IF_4(x2060, x2061, x2059, x2062, x2063, x2058, empty(fstsplit(x2058, app(map_f(two, head(x2061)), x2059))))) ==> IF_2(x1368, x1369, x1370, x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, x1370, x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), x1370))))) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (5) (length(x1370)=0 ==> IF_2(x1368, x1369, x1370, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, x1370, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), x1370))))) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on length(x1370)=s(x2057) which results in the following new constraint: (6) (s(length(x2066))=s(x2057) & leq(x2058, x2057)=false & (\/x2059,x2060,x2061,x2062,x2063:leq(x2058, x2057)=false & length(x2059)=x2057 ==> IF_2(x2060, x2061, x2059, x2062, x2063, x2058, false)_>=_IF_4(x2060, x2061, x2059, x2062, x2063, x2058, empty(fstsplit(x2058, app(map_f(two, head(x2061)), x2059))))) & (\/x2068,x2069,x2070,x2071,x2072,x2073,x2074,x2075,x2076,x2077,x2078:length(x2066)=s(x2068) & leq(x2069, x2068)=false & (\/x2070,x2071,x2072,x2073,x2074:leq(x2069, x2068)=false & length(x2070)=x2068 ==> IF_2(x2071, x2072, x2070, x2073, x2074, x2069, false)_>=_IF_4(x2071, x2072, x2070, x2073, x2074, x2069, empty(fstsplit(x2069, app(map_f(two, head(x2072)), x2070))))) ==> IF_2(x2075, x2076, x2066, x2077, x2078, s(x2069), false)_>=_IF_4(x2075, x2076, x2066, x2077, x2078, s(x2069), empty(fstsplit(s(x2069), app(map_f(two, head(x2076)), x2066))))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on length(x1370)=0 which results in the following new constraint: (7) (0=0 ==> IF_2(x1368, x1369, nil, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, nil, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), nil))))) We simplified constraint (7) using rules (I), (II) which results in the following new constraint: (8) (IF_2(x1368, x1369, nil, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, nil, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), nil))))) We simplified constraint (6) using rules (I), (II) which results in the following new constraint: (9) (length(x2066)=x2057 & leq(x2058, x2057)=false & (\/x2059,x2060,x2061,x2062,x2063:leq(x2058, x2057)=false & length(x2059)=x2057 ==> IF_2(x2060, x2061, x2059, x2062, x2063, x2058, false)_>=_IF_4(x2060, x2061, x2059, x2062, x2063, x2058, empty(fstsplit(x2058, app(map_f(two, head(x2061)), x2059))))) & (\/x2068,x2069,x2070,x2071,x2072,x2073,x2074,x2075,x2076,x2077,x2078:length(x2066)=s(x2068) & leq(x2069, x2068)=false & (\/x2070,x2071,x2072,x2073,x2074:leq(x2069, x2068)=false & length(x2070)=x2068 ==> IF_2(x2071, x2072, x2070, x2073, x2074, x2069, false)_>=_IF_4(x2071, x2072, x2070, x2073, x2074, x2069, empty(fstsplit(x2069, app(map_f(two, head(x2072)), x2070))))) ==> IF_2(x2075, x2076, x2066, x2077, x2078, s(x2069), false)_>=_IF_4(x2075, x2076, x2066, x2077, x2078, s(x2069), empty(fstsplit(s(x2069), app(map_f(two, head(x2076)), x2066))))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) We simplified constraint (9) using rule (VI) where we applied the induction hypothesis (\/x2059,x2060,x2061,x2062,x2063:leq(x2058, x2057)=false & length(x2059)=x2057 ==> IF_2(x2060, x2061, x2059, x2062, x2063, x2058, false)_>=_IF_4(x2060, x2061, x2059, x2062, x2063, x2058, empty(fstsplit(x2058, app(map_f(two, head(x2061)), x2059))))) with sigma = [x2059 / x2066, x2060 / x1368, x2061 / x1369, x2062 / x1371, x2063 / x1372] which results in the following new constraint: (10) (IF_2(x1368, x1369, x2066, x1371, x1372, x2058, false)_>=_IF_4(x1368, x1369, x2066, x1371, x1372, x2058, empty(fstsplit(x2058, app(map_f(two, head(x1369)), x2066)))) & (\/x2068,x2069,x2070,x2071,x2072,x2073,x2074,x2075,x2076,x2077,x2078:length(x2066)=s(x2068) & leq(x2069, x2068)=false & (\/x2070,x2071,x2072,x2073,x2074:leq(x2069, x2068)=false & length(x2070)=x2068 ==> IF_2(x2071, x2072, x2070, x2073, x2074, x2069, false)_>=_IF_4(x2071, x2072, x2070, x2073, x2074, x2069, empty(fstsplit(x2069, app(map_f(two, head(x2072)), x2070))))) ==> IF_2(x2075, x2076, x2066, x2077, x2078, s(x2069), false)_>=_IF_4(x2075, x2076, x2066, x2077, x2078, s(x2069), empty(fstsplit(s(x2069), app(map_f(two, head(x2076)), x2066))))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) We simplified constraint (10) using rule (IV) which results in the following new constraint: (11) (IF_2(x1368, x1369, x2066, x1371, x1372, x2058, false)_>=_IF_4(x1368, x1369, x2066, x1371, x1372, x2058, empty(fstsplit(x2058, app(map_f(two, head(x1369)), x2066)))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) For Pair IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) the following chains were created: *We consider the chain IF_2(x1532, x1533, x1534, x1535, x1536, x1537, false) -> IF_4(x1532, x1533, x1534, x1535, x1536, x1537, empty(fstsplit(x1537, app(map_f(two, head(x1533)), x1534)))), IF_4(x1538, x1539, x1540, x1541, x1542, x1543, false) -> RING(x1538, tail(x1539), sndsplit(x1543, app(map_f(two, head(x1539)), x1540)), cons(fstsplit(x1543, app(map_f(two, head(x1539)), x1540)), x1541), x1542, x1543) which results in the following constraint: (1) (IF_4(x1532, x1533, x1534, x1535, x1536, x1537, empty(fstsplit(x1537, app(map_f(two, head(x1533)), x1534))))=IF_4(x1538, x1539, x1540, x1541, x1542, x1543, false) ==> IF_4(x1538, x1539, x1540, x1541, x1542, x1543, false)_>=_RING(x1538, tail(x1539), sndsplit(x1543, app(map_f(two, head(x1539)), x1540)), cons(fstsplit(x1543, app(map_f(two, head(x1539)), x1540)), x1541), x1542, x1543)) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (fstsplit(x1537, x2080)=x2079 & empty(x2079)=false ==> IF_4(x1532, x1533, x1534, x1535, x1536, x1537, false)_>=_RING(x1532, tail(x1533), sndsplit(x1537, app(map_f(two, head(x1533)), x1534)), cons(fstsplit(x1537, app(map_f(two, head(x1533)), x1534)), x1535), x1536, x1537)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on empty(x2079)=false which results in the following new constraint: (3) (false=false & fstsplit(x1537, x2080)=cons(x2085, x2084) ==> IF_4(x1532, x1533, x1534, x1535, x1536, x1537, false)_>=_RING(x1532, tail(x1533), sndsplit(x1537, app(map_f(two, head(x1533)), x1534)), cons(fstsplit(x1537, app(map_f(two, head(x1533)), x1534)), x1535), x1536, x1537)) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (4) (fstsplit(x1537, x2080)=cons(x2085, x2084) ==> IF_4(x1532, x1533, x1534, x1535, x1536, x1537, false)_>=_RING(x1532, tail(x1533), sndsplit(x1537, app(map_f(two, head(x1533)), x1534)), cons(fstsplit(x1537, app(map_f(two, head(x1533)), x1534)), x1535), x1536, x1537)) We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on fstsplit(x1537, x2080)=cons(x2085, x2084) which results in the following new constraint: (5) (cons(x2089, fstsplit(x2090, x2088))=cons(x2085, x2084) & (\/x2091,x2092,x2093,x2094,x2095,x2096,x2097:fstsplit(x2090, x2088)=cons(x2091, x2092) ==> IF_4(x2093, x2094, x2095, x2096, x2097, x2090, false)_>=_RING(x2093, tail(x2094), sndsplit(x2090, app(map_f(two, head(x2094)), x2095)), cons(fstsplit(x2090, app(map_f(two, head(x2094)), x2095)), x2096), x2097, x2090)) ==> IF_4(x1532, x1533, x1534, x1535, x1536, s(x2090), false)_>=_RING(x1532, tail(x1533), sndsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), cons(fstsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), x1535), x1536, s(x2090))) We simplified constraint (5) using rules (I), (II), (IV) which results in the following new constraint: (6) (IF_4(x1532, x1533, x1534, x1535, x1536, s(x2090), false)_>=_RING(x1532, tail(x1533), sndsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), cons(fstsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), x1535), x1536, s(x2090))) For Pair RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) the following chains were created: *We consider the chain IF_3(x1742, x1743, x1744, x1745, x1746, x1747, false) -> RING(x1742, x1743, sndsplit(x1747, x1744), cons(fstsplit(x1747, x1744), x1745), x1746, x1747), RING(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754) -> IF_9(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754, empty(map_f(three, x1751))) which results in the following constraint: (1) (RING(x1742, x1743, sndsplit(x1747, x1744), cons(fstsplit(x1747, x1744), x1745), x1746, x1747)=RING(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754) ==> RING(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754)_>=_IF_9(x1748, x1749, x1750, cons(x1751, x1752), x1753, x1754, empty(map_f(three, x1751)))) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (RING(x1742, x1743, x1750, cons(x1751, x1745), x1746, x1747)_>=_IF_9(x1742, x1743, x1750, cons(x1751, x1745), x1746, x1747, empty(map_f(three, x1751)))) *We consider the chain IF_9(x1793, x1794, x1795, x1796, x1797, x1798, true) -> RING(x1793, x1794, x1795, tail(x1796), x1797, x1798), RING(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805) -> IF_9(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805, empty(map_f(three, x1802))) which results in the following constraint: (1) (RING(x1793, x1794, x1795, tail(x1796), x1797, x1798)=RING(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805) ==> RING(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805)_>=_IF_9(x1799, x1800, x1801, cons(x1802, x1803), x1804, x1805, empty(map_f(three, x1802)))) We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint: (2) (tail(x1796)=cons(x1802, x1803) ==> RING(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798)_>=_IF_9(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798, empty(map_f(three, x1802)))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on tail(x1796)=cons(x1802, x1803) which results in the following new constraint: (3) (x2110=cons(x1802, x1803) ==> RING(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798)_>=_IF_9(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798, empty(map_f(three, x1802)))) We simplified constraint (3) using rule (III) which results in the following new constraint: (4) (RING(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798)_>=_IF_9(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798, empty(map_f(three, x1802)))) *We consider the chain IF_8(x1812, x1813, x1814, x1815, x1816, x1817, false) -> RING(x1812, x1813, x1814, tail(x1815), sndsplit(x1817, app(map_f(three, head(x1815)), x1816)), x1817), RING(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824) -> IF_9(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824, empty(map_f(three, x1821))) which results in the following constraint: (1) (RING(x1812, x1813, x1814, tail(x1815), sndsplit(x1817, app(map_f(three, head(x1815)), x1816)), x1817)=RING(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824) ==> RING(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824)_>=_IF_9(x1818, x1819, x1820, cons(x1821, x1822), x1823, x1824, empty(map_f(three, x1821)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (tail(x1815)=cons(x1821, x1822) & app(x2113, x1816)=x2112 & sndsplit(x1817, x2112)=x1823 ==> RING(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817)_>=_IF_9(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817, empty(map_f(three, x1821)))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on tail(x1815)=cons(x1821, x1822) which results in the following new constraint: (3) (x2116=cons(x1821, x1822) & app(x2113, x1816)=x2112 & sndsplit(x1817, x2112)=x1823 ==> RING(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817)_>=_IF_9(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817, empty(map_f(three, x1821)))) We simplified constraint (3) using rules (III), (IV) which results in the following new constraint: (4) (RING(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817)_>=_IF_9(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817, empty(map_f(three, x1821)))) *We consider the chain IF_4(x1831, x1832, x1833, x1834, x1835, x1836, false) -> RING(x1831, tail(x1832), sndsplit(x1836, app(map_f(two, head(x1832)), x1833)), cons(fstsplit(x1836, app(map_f(two, head(x1832)), x1833)), x1834), x1835, x1836), RING(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843) -> IF_9(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843, empty(map_f(three, x1840))) which results in the following constraint: (1) (RING(x1831, tail(x1832), sndsplit(x1836, app(map_f(two, head(x1832)), x1833)), cons(fstsplit(x1836, app(map_f(two, head(x1832)), x1833)), x1834), x1835, x1836)=RING(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843) ==> RING(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843)_>=_IF_9(x1837, x1838, x1839, cons(x1840, x1841), x1842, x1843, empty(map_f(three, x1840)))) We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint: (2) (RING(x1831, x1838, x1839, cons(x1840, x1834), x1835, x1836)_>=_IF_9(x1831, x1838, x1839, cons(x1840, x1834), x1835, x1836, empty(map_f(three, x1840)))) To summarize, we get the following constraints P__>=_ for the following pairs. *RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) *(RING(x274, x275, x282, x283, x278, x279)_>=_IF_2(x274, x275, x282, x283, x278, x279, leq(x279, length(x282)))) *(RING(x322, x323, x324, x331, x326, x327)_>=_IF_2(x322, x323, x324, x331, x326, x327, leq(x327, length(x324)))) *(RING(x340, x341, x342, x349, x350, x345)_>=_IF_2(x340, x341, x342, x349, x350, x345, leq(x345, length(x342)))) *(RING(x358, x365, x366, x367, x362, x363)_>=_IF_2(x358, x365, x366, x367, x362, x363, leq(x363, length(x366)))) *IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) *(IF_2(x396, x397, x398, x399, x400, 0, true)_>=_IF_3(x396, x397, x398, x399, x400, 0, empty(fstsplit(0, x398)))) *(IF_2(x396, x397, x1898, x399, x400, x1892, true)_>=_IF_3(x396, x397, x1898, x399, x400, x1892, empty(fstsplit(x1892, x1898))) ==> IF_2(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), true)_>=_IF_3(x396, x397, cons(x1899, x1898), x399, x400, s(x1892), empty(fstsplit(s(x1892), cons(x1899, x1898))))) *IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) *(IF_3(x506, x507, cons(x1917, x1916), x509, x510, s(x1918), false)_>=_RING(x506, x507, sndsplit(s(x1918), cons(x1917, x1916)), cons(fstsplit(s(x1918), cons(x1917, x1916)), x509), x510, s(x1918))) *RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) *(RING(x726, x727, x734, x735, x730, x731)_>=_IF_6(x726, x727, x734, x735, x730, x731, leq(x731, length(x730)))) *(RING(x774, x775, x776, x783, x778, x779)_>=_IF_6(x774, x775, x776, x783, x778, x779, leq(x779, length(x778)))) *(RING(x792, x793, x794, x801, x802, x797)_>=_IF_6(x792, x793, x794, x801, x802, x797, leq(x797, length(x802)))) *(RING(x810, x817, x818, x819, x814, x815)_>=_IF_6(x810, x817, x818, x819, x814, x815, leq(x815, length(x814)))) *IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) *(IF_9(x1135, x1136, x1137, cons(nil, x1139), x1140, x1141, true)_>=_RING(x1135, x1136, x1137, tail(cons(nil, x1139)), x1140, x1141)) *IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) *(IF_6(x1184, x1185, x1186, x1187, nil, s(x2012), false)_>=_IF_8(x1184, x1185, x1186, x1187, nil, s(x2012), empty(fstsplit(s(x2012), app(map_f(three, head(x1187)), nil))))) *(IF_6(x1184, x1185, x1186, x1187, x2022, x2014, false)_>=_IF_8(x1184, x1185, x1186, x1187, x2022, x2014, empty(fstsplit(x2014, app(map_f(three, head(x1187)), x2022)))) ==> IF_6(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), false)_>=_IF_8(x1184, x1185, x1186, x1187, cons(x2023, x2022), s(x2014), empty(fstsplit(s(x2014), app(map_f(three, head(x1187)), cons(x2023, x2022)))))) *IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) *(IF_8(x1312, x1313, x1314, x1315, x1316, s(x2046), false)_>=_RING(x1312, x1313, x1314, tail(x1315), sndsplit(s(x2046), app(map_f(three, head(x1315)), x1316)), s(x2046))) *IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) *(IF_2(x1368, x1369, nil, x1371, x1372, s(x2056), false)_>=_IF_4(x1368, x1369, nil, x1371, x1372, s(x2056), empty(fstsplit(s(x2056), app(map_f(two, head(x1369)), nil))))) *(IF_2(x1368, x1369, x2066, x1371, x1372, x2058, false)_>=_IF_4(x1368, x1369, x2066, x1371, x1372, x2058, empty(fstsplit(x2058, app(map_f(two, head(x1369)), x2066)))) ==> IF_2(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), false)_>=_IF_4(x1368, x1369, cons(x2067, x2066), x1371, x1372, s(x2058), empty(fstsplit(s(x2058), app(map_f(two, head(x1369)), cons(x2067, x2066)))))) *IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) *(IF_4(x1532, x1533, x1534, x1535, x1536, s(x2090), false)_>=_RING(x1532, tail(x1533), sndsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), cons(fstsplit(s(x2090), app(map_f(two, head(x1533)), x1534)), x1535), x1536, s(x2090))) *RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) *(RING(x1742, x1743, x1750, cons(x1751, x1745), x1746, x1747)_>=_IF_9(x1742, x1743, x1750, cons(x1751, x1745), x1746, x1747, empty(map_f(three, x1751)))) *(RING(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798)_>=_IF_9(x1793, x1794, x1795, cons(x1802, x1803), x1797, x1798, empty(map_f(three, x1802)))) *(RING(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817)_>=_IF_9(x1812, x1813, x1814, cons(x1821, x1822), x1823, x1817, empty(map_f(three, x1821)))) *(RING(x1831, x1838, x1839, cons(x1840, x1834), x1835, x1836)_>=_IF_9(x1831, x1838, x1839, cons(x1840, x1834), x1835, x1836, empty(map_f(three, x1840)))) The constraints for P_> respective P_bound are constructed from P__>=_ where we just replace every occurence of "t _>=_ s" in P__>=_ by "t > s" respective "t _>=_ c". Here c stands for the fresh constant used for P_bound. Using the following integer polynomial ordering the resulting constraints can be solved Polynomial interpretation [NONINF]: POL(0) = 0 POL(IF_2(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 - x_3 + x_7 POL(IF_3(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = -1 + x_1 - x_3 POL(IF_4(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = -1 + x_1 POL(IF_6(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 - x_3 + x_7 POL(IF_8(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 - x_3 + x_7 POL(IF_9(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 - x_3 - x_4 + x_7 POL(RING(x_1, x_2, x_3, x_4, x_5, x_6)) = x_1 - x_3 POL(app(x_1, x_2)) = 0 POL(c) = -1 POL(cons(x_1, x_2)) = 0 POL(empty(x_1)) = 0 POL(f(x_1, x_2)) = x_1 + x_2 POL(false) = 0 POL(fstsplit(x_1, x_2)) = 0 POL(head(x_1)) = 0 POL(length(x_1)) = 0 POL(leq(x_1, x_2)) = 0 POL(map_f(x_1, x_2)) = 0 POL(nil) = 1 POL(s(x_1)) = 1 POL(sndsplit(x_1, x_2)) = x_1 POL(tail(x_1)) = 0 POL(three) = 0 POL(true) = 0 POL(two) = 0 The following pairs are in P_>: IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) The following pairs are in P_bound: IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) The following rules are usable: empty(nil) -> true empty(cons(h, t)) -> false x -> sndsplit(0, x) nil -> sndsplit(s(n), nil) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) ---------------------------------------- (71) Complex Obligation (AND) ---------------------------------------- (72) Obligation: Q DP problem: The TRS P consists of the following rules: IF_3(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (73) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (74) Obligation: Q DP problem: The TRS P consists of the following rules: IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (75) QDPQMonotonicMRRProof (EQUIVALENT) By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain. Strictly oriented rules of the TRS R: leq(0, m) -> true Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(IF_2(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 + x_7 POL(IF_4(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = 1 + x_1 POL(IF_6(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_1 + x_7 POL(IF_8(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = 1 + x_1 POL(IF_9(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = 1 + x_1 POL(RING(x_1, x_2, x_3, x_4, x_5, x_6)) = 1 + x_1 POL(app(x_1, x_2)) = 2*x_1 + 2*x_2 POL(cons(x_1, x_2)) = 1 + 2*x_1 + x_2 POL(empty(x_1)) = 2 POL(f(x_1, x_2)) = 0 POL(false) = 1 POL(fstsplit(x_1, x_2)) = 2*x_2 POL(head(x_1)) = x_1 POL(length(x_1)) = x_1 POL(leq(x_1, x_2)) = 1 POL(map_f(x_1, x_2)) = 0 POL(nil) = 0 POL(s(x_1)) = 0 POL(sndsplit(x_1, x_2)) = 2 + x_2 POL(tail(x_1)) = 2*x_1 POL(three) = 0 POL(true) = 0 POL(two) = 0 ---------------------------------------- (76) Obligation: Q DP problem: The TRS P consists of the following rules: IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (77) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule IF_2(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2)))) at position [6] we obtained the following new rules [LPAR04]: (IF_2(y0, y1, y2, y3, y4, 0, false) -> IF_4(y0, y1, y2, y3, y4, 0, empty(nil)),IF_2(y0, y1, y2, y3, y4, 0, false) -> IF_4(y0, y1, y2, y3, y4, 0, empty(nil))) (IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2)))),IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2))))) ---------------------------------------- (78) Obligation: Q DP problem: The TRS P consists of the following rules: IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) IF_2(y0, y1, y2, y3, y4, 0, false) -> IF_4(y0, y1, y2, y3, y4, 0, empty(nil)) IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2)))) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (79) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (80) Obligation: Q DP problem: The TRS P consists of the following rules: RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (81) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) at position [6] we obtained the following new rules [LPAR04]: (RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)),RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0))) (RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))),RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1))))) ---------------------------------------- (82) Obligation: Q DP problem: The TRS P consists of the following rules: IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (83) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) at position [6] we obtained the following new rules [LPAR04]: (RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)),RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0))) (RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))),RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1))))) ---------------------------------------- (84) Obligation: Q DP problem: The TRS P consists of the following rules: IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (85) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) at position [6] we obtained the following new rules [LPAR04]: (IF_6(y0, y1, y2, y3, y4, 0, false) -> IF_8(y0, y1, y2, y3, y4, 0, empty(nil)),IF_6(y0, y1, y2, y3, y4, 0, false) -> IF_8(y0, y1, y2, y3, y4, 0, empty(nil))) (IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))),IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4))))) ---------------------------------------- (86) Obligation: Q DP problem: The TRS P consists of the following rules: IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2)))) IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_6(y0, y1, y2, y3, y4, 0, false) -> IF_8(y0, y1, y2, y3, y4, 0, empty(nil)) IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (87) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (88) Obligation: Q DP problem: The TRS P consists of the following rules: IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2)))) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (89) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule IF_4(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m) we obtained the following new rules [LPAR04]: (IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, tail(cons(z1, z2)), sndsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), cons(fstsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), z4), z5, z6),IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, tail(cons(z1, z2)), sndsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), cons(fstsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), z4), z5, z6)) ---------------------------------------- (90) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2)))) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, tail(cons(z1, z2)), sndsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), cons(fstsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), z4), z5, z6) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (91) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, tail(cons(z1, z2)), sndsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), cons(fstsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), z4), z5, z6) at position [1] we obtained the following new rules [LPAR04]: (IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), cons(fstsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), z4), z5, z6),IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), cons(fstsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), z4), z5, z6)) ---------------------------------------- (92) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2)))) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), cons(fstsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), z4), z5, z6) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (93) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), cons(fstsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), z4), z5, z6) at position [2,1,0,1] we obtained the following new rules [LPAR04]: (IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), z4), z5, z6),IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), z4), z5, z6)) ---------------------------------------- (94) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2)))) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), z4), z5, z6) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (95) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, head(cons(z1, z2))), z3)), z4), z5, z6) at position [3,0,1,0,1] we obtained the following new rules [LPAR04]: (IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6),IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6)) ---------------------------------------- (96) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2)))) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (97) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) we obtained the following new rules [LPAR04]: (IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, tail(cons(z3, z4)), z5, z6),IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, tail(cons(z3, z4)), z5, z6)) ---------------------------------------- (98) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2)))) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, tail(cons(z3, z4)), z5, z6) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (99) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, tail(cons(z3, z4)), z5, z6) at position [3] we obtained the following new rules [LPAR04]: (IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6),IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6)) ---------------------------------------- (100) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2)))) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (101) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule IF_2(y0, cons(x0, x1), y2, y3, y4, y5, false) -> IF_4(y0, cons(x0, x1), y2, y3, y4, y5, empty(fstsplit(y5, app(map_f(two, x0), y2)))) we obtained the following new rules [LPAR04]: (IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))),IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil))))) (IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))),IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3)))))) ---------------------------------------- (102) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (103) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) we obtained the following new rules [LPAR04]: (IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))),IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil))))) (IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))),IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5)))))) ---------------------------------------- (104) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (105) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) we obtained the following new rules [LPAR04]: (IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, tail(cons(z3, z4)), sndsplit(z5, app(map_f(three, head(cons(z3, z4))), nil)), z5),IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, tail(cons(z3, z4)), sndsplit(z5, app(map_f(three, head(cons(z3, z4))), nil)), z5)) (IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, tail(cons(z3, z4)), sndsplit(z7, app(map_f(three, head(cons(z3, z4))), cons(z5, z6))), z7),IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, tail(cons(z3, z4)), sndsplit(z7, app(map_f(three, head(cons(z3, z4))), cons(z5, z6))), z7)) ---------------------------------------- (106) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, tail(cons(z3, z4)), sndsplit(z5, app(map_f(three, head(cons(z3, z4))), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, tail(cons(z3, z4)), sndsplit(z7, app(map_f(three, head(cons(z3, z4))), cons(z5, z6))), z7) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (107) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, tail(cons(z3, z4)), sndsplit(z5, app(map_f(three, head(cons(z3, z4))), nil)), z5) at position [3] we obtained the following new rules [LPAR04]: (IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, head(cons(z3, z4))), nil)), z5),IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, head(cons(z3, z4))), nil)), z5)) ---------------------------------------- (108) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, tail(cons(z3, z4)), sndsplit(z7, app(map_f(three, head(cons(z3, z4))), cons(z5, z6))), z7) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, head(cons(z3, z4))), nil)), z5) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (109) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, tail(cons(z3, z4)), sndsplit(z7, app(map_f(three, head(cons(z3, z4))), cons(z5, z6))), z7) at position [3] we obtained the following new rules [LPAR04]: (IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, head(cons(z3, z4))), cons(z5, z6))), z7),IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, head(cons(z3, z4))), cons(z5, z6))), z7)) ---------------------------------------- (110) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, head(cons(z3, z4))), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, head(cons(z3, z4))), cons(z5, z6))), z7) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (111) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (112) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, head(cons(z3, z4))), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, head(cons(z3, z4))), cons(z5, z6))), z7) The TRS R consists of the following rules: head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (113) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. tail(cons(x0, x1)) ---------------------------------------- (114) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, head(cons(z3, z4))), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, head(cons(z3, z4))), cons(z5, z6))), z7) The TRS R consists of the following rules: head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (115) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, head(cons(z3, z4))), nil)), z5) at position [4,1,0,1] we obtained the following new rules [LPAR04]: (IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5),IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5)) ---------------------------------------- (116) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, head(cons(z3, z4))), cons(z5, z6))), z7) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) The TRS R consists of the following rules: head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (117) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, head(cons(z3, z4))), cons(z5, z6))), z7) at position [4,1,0,1] we obtained the following new rules [LPAR04]: (IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7),IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7)) ---------------------------------------- (118) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) The TRS R consists of the following rules: head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (119) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (120) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (121) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. head(cons(x0, x1)) ---------------------------------------- (122) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (123) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule IF_4(z0, cons(z1, z2), z3, z4, z5, z6, false) -> RING(z0, z2, sndsplit(z6, app(map_f(two, z1), z3)), cons(fstsplit(z6, app(map_f(two, z1), z3)), z4), z5, z6) we obtained the following new rules [LPAR04]: (IF_4(z0, cons(z1, z2), nil, z3, z4, z5, false) -> RING(z0, z2, sndsplit(z5, app(map_f(two, z1), nil)), cons(fstsplit(z5, app(map_f(two, z1), nil)), z3), z4, z5),IF_4(z0, cons(z1, z2), nil, z3, z4, z5, false) -> RING(z0, z2, sndsplit(z5, app(map_f(two, z1), nil)), cons(fstsplit(z5, app(map_f(two, z1), nil)), z3), z4, z5)) (IF_4(z0, cons(z1, z2), cons(z3, z4), z5, z6, z7, false) -> RING(z0, z2, sndsplit(z7, app(map_f(two, z1), cons(z3, z4))), cons(fstsplit(z7, app(map_f(two, z1), cons(z3, z4))), z5), z6, z7),IF_4(z0, cons(z1, z2), cons(z3, z4), z5, z6, z7, false) -> RING(z0, z2, sndsplit(z7, app(map_f(two, z1), cons(z3, z4))), cons(fstsplit(z7, app(map_f(two, z1), cons(z3, z4))), z5), z6, z7)) ---------------------------------------- (124) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) IF_4(z0, cons(z1, z2), nil, z3, z4, z5, false) -> RING(z0, z2, sndsplit(z5, app(map_f(two, z1), nil)), cons(fstsplit(z5, app(map_f(two, z1), nil)), z3), z4, z5) IF_4(z0, cons(z1, z2), cons(z3, z4), z5, z6, z7, false) -> RING(z0, z2, sndsplit(z7, app(map_f(two, z1), cons(z3, z4))), cons(fstsplit(z7, app(map_f(two, z1), cons(z3, z4))), z5), z6, z7) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (125) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule RING(y0, y1, nil, y3, y4, y5) -> IF_2(y0, y1, nil, y3, y4, y5, leq(y5, 0)) we obtained the following new rules [LPAR04]: (RING(x0, cons(y_1, y_2), nil, x2, x3, x4) -> IF_2(x0, cons(y_1, y_2), nil, x2, x3, x4, leq(x4, 0)),RING(x0, cons(y_1, y_2), nil, x2, x3, x4) -> IF_2(x0, cons(y_1, y_2), nil, x2, x3, x4, leq(x4, 0))) ---------------------------------------- (126) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) IF_4(z0, cons(z1, z2), nil, z3, z4, z5, false) -> RING(z0, z2, sndsplit(z5, app(map_f(two, z1), nil)), cons(fstsplit(z5, app(map_f(two, z1), nil)), z3), z4, z5) IF_4(z0, cons(z1, z2), cons(z3, z4), z5, z6, z7, false) -> RING(z0, z2, sndsplit(z7, app(map_f(two, z1), cons(z3, z4))), cons(fstsplit(z7, app(map_f(two, z1), cons(z3, z4))), z5), z6, z7) RING(x0, cons(y_1, y_2), nil, x2, x3, x4) -> IF_2(x0, cons(y_1, y_2), nil, x2, x3, x4, leq(x4, 0)) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (127) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule RING(y0, y1, cons(x0, x1), y3, y4, y5) -> IF_2(y0, y1, cons(x0, x1), y3, y4, y5, leq(y5, s(length(x1)))) we obtained the following new rules [LPAR04]: (RING(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6) -> IF_2(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6, leq(x6, s(length(x3)))),RING(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6) -> IF_2(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6, leq(x6, s(length(x3))))) ---------------------------------------- (128) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) IF_4(z0, cons(z1, z2), nil, z3, z4, z5, false) -> RING(z0, z2, sndsplit(z5, app(map_f(two, z1), nil)), cons(fstsplit(z5, app(map_f(two, z1), nil)), z3), z4, z5) IF_4(z0, cons(z1, z2), cons(z3, z4), z5, z6, z7, false) -> RING(z0, z2, sndsplit(z7, app(map_f(two, z1), cons(z3, z4))), cons(fstsplit(z7, app(map_f(two, z1), cons(z3, z4))), z5), z6, z7) RING(x0, cons(y_1, y_2), nil, x2, x3, x4) -> IF_2(x0, cons(y_1, y_2), nil, x2, x3, x4, leq(x4, 0)) RING(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6) -> IF_2(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6, leq(x6, s(length(x3)))) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (129) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) we obtained the following new rules [LPAR04]: (RING(x0, x1, x2, cons(y_3, y_4), nil, x4) -> IF_6(x0, x1, x2, cons(y_3, y_4), nil, x4, leq(x4, 0)),RING(x0, x1, x2, cons(y_3, y_4), nil, x4) -> IF_6(x0, x1, x2, cons(y_3, y_4), nil, x4, leq(x4, 0))) ---------------------------------------- (130) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) IF_4(z0, cons(z1, z2), nil, z3, z4, z5, false) -> RING(z0, z2, sndsplit(z5, app(map_f(two, z1), nil)), cons(fstsplit(z5, app(map_f(two, z1), nil)), z3), z4, z5) IF_4(z0, cons(z1, z2), cons(z3, z4), z5, z6, z7, false) -> RING(z0, z2, sndsplit(z7, app(map_f(two, z1), cons(z3, z4))), cons(fstsplit(z7, app(map_f(two, z1), cons(z3, z4))), z5), z6, z7) RING(x0, cons(y_1, y_2), nil, x2, x3, x4) -> IF_2(x0, cons(y_1, y_2), nil, x2, x3, x4, leq(x4, 0)) RING(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6) -> IF_2(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6, leq(x6, s(length(x3)))) RING(x0, x1, x2, cons(y_3, y_4), nil, x4) -> IF_6(x0, x1, x2, cons(y_3, y_4), nil, x4, leq(x4, 0)) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (131) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) we obtained the following new rules [LPAR04]: (RING(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6) -> IF_6(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6, leq(x6, s(length(x5)))),RING(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6) -> IF_6(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6, leq(x6, s(length(x5))))) ---------------------------------------- (132) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) IF_4(z0, cons(z1, z2), nil, z3, z4, z5, false) -> RING(z0, z2, sndsplit(z5, app(map_f(two, z1), nil)), cons(fstsplit(z5, app(map_f(two, z1), nil)), z3), z4, z5) IF_4(z0, cons(z1, z2), cons(z3, z4), z5, z6, z7, false) -> RING(z0, z2, sndsplit(z7, app(map_f(two, z1), cons(z3, z4))), cons(fstsplit(z7, app(map_f(two, z1), cons(z3, z4))), z5), z6, z7) RING(x0, cons(y_1, y_2), nil, x2, x3, x4) -> IF_2(x0, cons(y_1, y_2), nil, x2, x3, x4, leq(x4, 0)) RING(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6) -> IF_2(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6, leq(x6, s(length(x3)))) RING(x0, x1, x2, cons(y_3, y_4), nil, x4) -> IF_6(x0, x1, x2, cons(y_3, y_4), nil, x4, leq(x4, 0)) RING(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6) -> IF_6(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6, leq(x6, s(length(x5)))) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (133) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule IF_9(z0, z1, z2, cons(z3, z4), z5, z6, true) -> RING(z0, z1, z2, z4, z5, z6) we obtained the following new rules [LPAR04]: (IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), x5, x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), x5, x6),IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), x5, x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), x5, x6)) (IF_9(x0, cons(y_1, y_2), nil, cons(x3, x4), x5, x6, true) -> RING(x0, cons(y_1, y_2), nil, x4, x5, x6),IF_9(x0, cons(y_1, y_2), nil, cons(x3, x4), x5, x6, true) -> RING(x0, cons(y_1, y_2), nil, x4, x5, x6)) (IF_9(x0, cons(y_1, y_2), cons(y_3, y_4), cons(x3, x4), x5, x6, true) -> RING(x0, cons(y_1, y_2), cons(y_3, y_4), x4, x5, x6),IF_9(x0, cons(y_1, y_2), cons(y_3, y_4), cons(x3, x4), x5, x6, true) -> RING(x0, cons(y_1, y_2), cons(y_3, y_4), x4, x5, x6)) (IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), nil, x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), nil, x6),IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), nil, x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), nil, x6)) (IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), cons(y_5, y_6), x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), cons(y_5, y_6), x6),IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), cons(y_5, y_6), x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), cons(y_5, y_6), x6)) ---------------------------------------- (134) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) IF_4(z0, cons(z1, z2), nil, z3, z4, z5, false) -> RING(z0, z2, sndsplit(z5, app(map_f(two, z1), nil)), cons(fstsplit(z5, app(map_f(two, z1), nil)), z3), z4, z5) IF_4(z0, cons(z1, z2), cons(z3, z4), z5, z6, z7, false) -> RING(z0, z2, sndsplit(z7, app(map_f(two, z1), cons(z3, z4))), cons(fstsplit(z7, app(map_f(two, z1), cons(z3, z4))), z5), z6, z7) RING(x0, cons(y_1, y_2), nil, x2, x3, x4) -> IF_2(x0, cons(y_1, y_2), nil, x2, x3, x4, leq(x4, 0)) RING(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6) -> IF_2(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6, leq(x6, s(length(x3)))) RING(x0, x1, x2, cons(y_3, y_4), nil, x4) -> IF_6(x0, x1, x2, cons(y_3, y_4), nil, x4, leq(x4, 0)) RING(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6) -> IF_6(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6, leq(x6, s(length(x5)))) IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), x5, x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), x5, x6) IF_9(x0, cons(y_1, y_2), nil, cons(x3, x4), x5, x6, true) -> RING(x0, cons(y_1, y_2), nil, x4, x5, x6) IF_9(x0, cons(y_1, y_2), cons(y_3, y_4), cons(x3, x4), x5, x6, true) -> RING(x0, cons(y_1, y_2), cons(y_3, y_4), x4, x5, x6) IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), nil, x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), nil, x6) IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), cons(y_5, y_6), x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), cons(y_5, y_6), x6) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (135) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) we obtained the following new rules [LPAR04]: (RING(x0, x1, x2, cons(x3, cons(y_4, y_5)), x5, x6) -> IF_9(x0, x1, x2, cons(x3, cons(y_4, y_5)), x5, x6, empty(map_f(three, x3))),RING(x0, x1, x2, cons(x3, cons(y_4, y_5)), x5, x6) -> IF_9(x0, x1, x2, cons(x3, cons(y_4, y_5)), x5, x6, empty(map_f(three, x3)))) (RING(x0, cons(y_1, y_2), x2, cons(x3, x4), x5, x6) -> IF_9(x0, cons(y_1, y_2), x2, cons(x3, x4), x5, x6, empty(map_f(three, x3))),RING(x0, cons(y_1, y_2), x2, cons(x3, x4), x5, x6) -> IF_9(x0, cons(y_1, y_2), x2, cons(x3, x4), x5, x6, empty(map_f(three, x3)))) (RING(x0, cons(y_1, y_2), cons(y_3, y_4), cons(x3, x4), x5, x6) -> IF_9(x0, cons(y_1, y_2), cons(y_3, y_4), cons(x3, x4), x5, x6, empty(map_f(three, x3))),RING(x0, cons(y_1, y_2), cons(y_3, y_4), cons(x3, x4), x5, x6) -> IF_9(x0, cons(y_1, y_2), cons(y_3, y_4), cons(x3, x4), x5, x6, empty(map_f(three, x3)))) (RING(x0, x1, x2, cons(x3, cons(y_4, y_5)), cons(y_6, y_7), x6) -> IF_9(x0, x1, x2, cons(x3, cons(y_4, y_5)), cons(y_6, y_7), x6, empty(map_f(three, x3))),RING(x0, x1, x2, cons(x3, cons(y_4, y_5)), cons(y_6, y_7), x6) -> IF_9(x0, x1, x2, cons(x3, cons(y_4, y_5)), cons(y_6, y_7), x6, empty(map_f(three, x3)))) ---------------------------------------- (136) Obligation: Q DP problem: The TRS P consists of the following rules: IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) IF_4(z0, cons(z1, z2), nil, z3, z4, z5, false) -> RING(z0, z2, sndsplit(z5, app(map_f(two, z1), nil)), cons(fstsplit(z5, app(map_f(two, z1), nil)), z3), z4, z5) IF_4(z0, cons(z1, z2), cons(z3, z4), z5, z6, z7, false) -> RING(z0, z2, sndsplit(z7, app(map_f(two, z1), cons(z3, z4))), cons(fstsplit(z7, app(map_f(two, z1), cons(z3, z4))), z5), z6, z7) RING(x0, cons(y_1, y_2), nil, x2, x3, x4) -> IF_2(x0, cons(y_1, y_2), nil, x2, x3, x4, leq(x4, 0)) RING(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6) -> IF_2(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6, leq(x6, s(length(x3)))) RING(x0, x1, x2, cons(y_3, y_4), nil, x4) -> IF_6(x0, x1, x2, cons(y_3, y_4), nil, x4, leq(x4, 0)) RING(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6) -> IF_6(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6, leq(x6, s(length(x5)))) IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), x5, x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), x5, x6) IF_9(x0, cons(y_1, y_2), nil, cons(x3, x4), x5, x6, true) -> RING(x0, cons(y_1, y_2), nil, x4, x5, x6) IF_9(x0, cons(y_1, y_2), cons(y_3, y_4), cons(x3, x4), x5, x6, true) -> RING(x0, cons(y_1, y_2), cons(y_3, y_4), x4, x5, x6) IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), nil, x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), nil, x6) IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), cons(y_5, y_6), x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), cons(y_5, y_6), x6) RING(x0, x1, x2, cons(x3, cons(y_4, y_5)), x5, x6) -> IF_9(x0, x1, x2, cons(x3, cons(y_4, y_5)), x5, x6, empty(map_f(three, x3))) RING(x0, cons(y_1, y_2), x2, cons(x3, x4), x5, x6) -> IF_9(x0, cons(y_1, y_2), x2, cons(x3, x4), x5, x6, empty(map_f(three, x3))) RING(x0, cons(y_1, y_2), cons(y_3, y_4), cons(x3, x4), x5, x6) -> IF_9(x0, cons(y_1, y_2), cons(y_3, y_4), cons(x3, x4), x5, x6, empty(map_f(three, x3))) RING(x0, x1, x2, cons(x3, cons(y_4, y_5)), cons(y_6, y_7), x6) -> IF_9(x0, x1, x2, cons(x3, cons(y_4, y_5)), cons(y_6, y_7), x6, empty(map_f(three, x3))) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (137) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *IF_4(z0, cons(z1, z2), nil, z3, z4, z5, false) -> RING(z0, z2, sndsplit(z5, app(map_f(two, z1), nil)), cons(fstsplit(z5, app(map_f(two, z1), nil)), z3), z4, z5) The graph contains the following edges 1 >= 1, 2 > 2, 5 >= 5, 6 >= 6 *RING(x0, cons(y_1, y_2), nil, x2, x3, x4) -> IF_2(x0, cons(y_1, y_2), nil, x2, x3, x4, leq(x4, 0)) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *IF_4(z0, cons(z1, z2), cons(z3, z4), z5, z6, z7, false) -> RING(z0, z2, sndsplit(z7, app(map_f(two, z1), cons(z3, z4))), cons(fstsplit(z7, app(map_f(two, z1), cons(z3, z4))), z5), z6, z7) The graph contains the following edges 1 >= 1, 2 > 2, 5 >= 5, 6 >= 6 *RING(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6) -> IF_2(x0, cons(y_1, y_2), cons(x2, x3), x4, x5, x6, leq(x6, s(length(x3)))) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 > 4, 6 >= 6 *RING(x0, x1, x2, cons(y_3, y_4), nil, x4) -> IF_6(x0, x1, x2, cons(y_3, y_4), nil, x4, leq(x4, 0)) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 > 4, 6 >= 6 *RING(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6) -> IF_6(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6, leq(x6, s(length(x5)))) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *IF_2(z0, cons(x1, x2), nil, z2, z3, z4, false) -> IF_4(z0, cons(x1, x2), nil, z2, z3, z4, empty(fstsplit(z4, app(map_f(two, x1), nil)))) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *IF_2(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, false) -> IF_4(z0, cons(x1, x2), cons(z2, z3), z4, z5, z6, empty(fstsplit(z6, app(map_f(two, x1), cons(z2, z3))))) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *IF_9(x0, cons(y_1, y_2), cons(y_3, y_4), cons(x3, x4), x5, x6, true) -> RING(x0, cons(y_1, y_2), cons(y_3, y_4), x4, x5, x6) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 > 4, 5 >= 5, 6 >= 6 *IF_9(x0, cons(y_1, y_2), nil, cons(x3, x4), x5, x6, true) -> RING(x0, cons(y_1, y_2), nil, x4, x5, x6) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 > 4, 5 >= 5, 6 >= 6 *IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), cons(y_5, y_6), x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), cons(y_5, y_6), x6) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 > 4, 5 >= 5, 6 >= 6 *IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), x5, x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), x5, x6) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 > 4, 5 >= 5, 6 >= 6 *IF_9(x0, x1, x2, cons(x3, cons(y_3, y_4)), nil, x6, true) -> RING(x0, x1, x2, cons(y_3, y_4), nil, x6) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 > 4, 5 >= 5, 6 >= 6 *RING(x0, cons(y_1, y_2), cons(y_3, y_4), cons(x3, x4), x5, x6) -> IF_9(x0, cons(y_1, y_2), cons(y_3, y_4), cons(x3, x4), x5, x6, empty(map_f(three, x3))) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *RING(x0, x1, x2, cons(x3, cons(y_4, y_5)), cons(y_6, y_7), x6) -> IF_9(x0, x1, x2, cons(x3, cons(y_4, y_5)), cons(y_6, y_7), x6, empty(map_f(three, x3))) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *RING(x0, x1, x2, cons(x3, cons(y_4, y_5)), x5, x6) -> IF_9(x0, x1, x2, cons(x3, cons(y_4, y_5)), x5, x6, empty(map_f(three, x3))) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *RING(x0, cons(y_1, y_2), x2, cons(x3, x4), x5, x6) -> IF_9(x0, cons(y_1, y_2), x2, cons(x3, x4), x5, x6, empty(map_f(three, x3))) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 ---------------------------------------- (138) YES ---------------------------------------- (139) Obligation: Q DP problem: The TRS P consists of the following rules: IF_2(st_1, in_2, st_2, in_3, st_3, m, true) -> IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2))) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2))) RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (140) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (141) Obligation: Q DP problem: The TRS P consists of the following rules: RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (142) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. RING(y0, y1, y2, cons(x0, x1), y4, y5) -> IF_9(y0, y1, y2, cons(x0, x1), y4, y5, empty(map_f(three, x0))) IF_9(st_1, in_2, st_2, in_3, st_3, m, true) -> RING(st_1, in_2, st_2, tail(in_3), st_3, m) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO,RATPOLO]: POL(0) = [1/4] POL(IF_6(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = x_4 POL(IF_8(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = [1/2]x_4 POL(IF_9(x_1, x_2, x_3, x_4, x_5, x_6, x_7)) = [1/4] + [1/2]x_4 POL(RING(x_1, x_2, x_3, x_4, x_5, x_6)) = x_4 POL(app(x_1, x_2)) = 0 POL(cons(x_1, x_2)) = [4] + [1/4]x_1 + [2]x_2 POL(empty(x_1)) = 0 POL(f(x_1, x_2)) = 0 POL(false) = 0 POL(fstsplit(x_1, x_2)) = 0 POL(head(x_1)) = 0 POL(length(x_1)) = [4] + [1/4]x_1 POL(leq(x_1, x_2)) = [1] POL(map_f(x_1, x_2)) = 0 POL(nil) = [2] POL(s(x_1)) = x_1 POL(sndsplit(x_1, x_2)) = 0 POL(tail(x_1)) = [1/2]x_1 POL(three) = 0 POL(true) = 0 The value of delta used in the strict ordering is 1/4. The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: tail(cons(h, t)) -> t ---------------------------------------- (143) Obligation: Q DP problem: The TRS P consists of the following rules: RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (144) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule RING(st_1, in_2, st_2, in_3, st_3, m) -> IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3))) at position [6] we obtained the following new rules [LPAR04]: (RING(y0, y1, y2, y3, y4, 0) -> IF_6(y0, y1, y2, y3, y4, 0, true),RING(y0, y1, y2, y3, y4, 0) -> IF_6(y0, y1, y2, y3, y4, 0, true)) (RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)),RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0))) (RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))),RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1))))) ---------------------------------------- (145) Obligation: Q DP problem: The TRS P consists of the following rules: IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, y3, y4, 0) -> IF_6(y0, y1, y2, y3, y4, 0, true) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (146) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (147) Obligation: Q DP problem: The TRS P consists of the following rules: IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (148) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule IF_8(st_1, in_2, st_2, in_3, st_3, m, false) -> RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m) at position [4] we obtained the following new rules [LPAR04]: (IF_8(y0, y1, y2, y3, y4, 0, false) -> RING(y0, y1, y2, tail(y3), app(map_f(three, head(y3)), y4), 0),IF_8(y0, y1, y2, y3, y4, 0, false) -> RING(y0, y1, y2, tail(y3), app(map_f(three, head(y3)), y4), 0)) (IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, tail(cons(x0, x1)), sndsplit(y5, app(map_f(three, x0), y4)), y5),IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, tail(cons(x0, x1)), sndsplit(y5, app(map_f(three, x0), y4)), y5)) ---------------------------------------- (149) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_8(y0, y1, y2, y3, y4, 0, false) -> RING(y0, y1, y2, tail(y3), app(map_f(three, head(y3)), y4), 0) IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, tail(cons(x0, x1)), sndsplit(y5, app(map_f(three, x0), y4)), y5) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (150) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, tail(cons(x0, x1)), sndsplit(y5, app(map_f(three, x0), y4)), y5) at position [3] we obtained the following new rules [LPAR04]: (IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, x1, sndsplit(y5, app(map_f(three, x0), y4)), y5),IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, x1, sndsplit(y5, app(map_f(three, x0), y4)), y5)) ---------------------------------------- (151) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_8(y0, y1, y2, y3, y4, 0, false) -> RING(y0, y1, y2, tail(y3), app(map_f(three, head(y3)), y4), 0) IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, x1, sndsplit(y5, app(map_f(three, x0), y4)), y5) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (152) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule IF_6(st_1, in_2, st_2, in_3, st_3, m, false) -> IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3)))) at position [6] we obtained the following new rules [LPAR04]: (IF_6(y0, y1, y2, y3, y4, 0, false) -> IF_8(y0, y1, y2, y3, y4, 0, empty(nil)),IF_6(y0, y1, y2, y3, y4, 0, false) -> IF_8(y0, y1, y2, y3, y4, 0, empty(nil))) (IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))),IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4))))) ---------------------------------------- (153) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_8(y0, y1, y2, y3, y4, 0, false) -> RING(y0, y1, y2, tail(y3), app(map_f(three, head(y3)), y4), 0) IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, x1, sndsplit(y5, app(map_f(three, x0), y4)), y5) IF_6(y0, y1, y2, y3, y4, 0, false) -> IF_8(y0, y1, y2, y3, y4, 0, empty(nil)) IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (154) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (155) Obligation: Q DP problem: The TRS P consists of the following rules: IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) IF_8(y0, y1, y2, y3, y4, 0, false) -> RING(y0, y1, y2, tail(y3), app(map_f(three, head(y3)), y4), 0) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, x1, sndsplit(y5, app(map_f(three, x0), y4)), y5) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (156) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule IF_8(y0, y1, y2, y3, y4, 0, false) -> RING(y0, y1, y2, tail(y3), app(map_f(three, head(y3)), y4), 0) at position [4] we obtained the following new rules [LPAR04]: (IF_8(y0, y1, y2, cons(x0, x1), y4, 0, false) -> RING(y0, y1, y2, tail(cons(x0, x1)), app(map_f(three, x0), y4), 0),IF_8(y0, y1, y2, cons(x0, x1), y4, 0, false) -> RING(y0, y1, y2, tail(cons(x0, x1)), app(map_f(three, x0), y4), 0)) ---------------------------------------- (157) Obligation: Q DP problem: The TRS P consists of the following rules: IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, x1, sndsplit(y5, app(map_f(three, x0), y4)), y5) IF_8(y0, y1, y2, cons(x0, x1), y4, 0, false) -> RING(y0, y1, y2, tail(cons(x0, x1)), app(map_f(three, x0), y4), 0) The TRS R consists of the following rules: tail(cons(h, t)) -> t head(cons(h, t)) -> h map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), 0) -> false leq(s(n), s(m)) -> leq(n, m) The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (158) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (159) Obligation: Q DP problem: The TRS P consists of the following rules: IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, x1, sndsplit(y5, app(map_f(three, x0), y4)), y5) IF_8(y0, y1, y2, cons(x0, x1), y4, 0, false) -> RING(y0, y1, y2, tail(cons(x0, x1)), app(map_f(three, x0), y4), 0) The TRS R consists of the following rules: tail(cons(h, t)) -> t map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) head(cons(x0, x1)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (160) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. head(cons(x0, x1)) ---------------------------------------- (161) Obligation: Q DP problem: The TRS P consists of the following rules: IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, x1, sndsplit(y5, app(map_f(three, x0), y4)), y5) IF_8(y0, y1, y2, cons(x0, x1), y4, 0, false) -> RING(y0, y1, y2, tail(cons(x0, x1)), app(map_f(three, x0), y4), 0) The TRS R consists of the following rules: tail(cons(h, t)) -> t map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (162) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule IF_8(y0, y1, y2, cons(x0, x1), y4, 0, false) -> RING(y0, y1, y2, tail(cons(x0, x1)), app(map_f(three, x0), y4), 0) at position [3] we obtained the following new rules [LPAR04]: (IF_8(y0, y1, y2, cons(x0, x1), y4, 0, false) -> RING(y0, y1, y2, x1, app(map_f(three, x0), y4), 0),IF_8(y0, y1, y2, cons(x0, x1), y4, 0, false) -> RING(y0, y1, y2, x1, app(map_f(three, x0), y4), 0)) ---------------------------------------- (163) Obligation: Q DP problem: The TRS P consists of the following rules: IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, x1, sndsplit(y5, app(map_f(three, x0), y4)), y5) IF_8(y0, y1, y2, cons(x0, x1), y4, 0, false) -> RING(y0, y1, y2, x1, app(map_f(three, x0), y4), 0) The TRS R consists of the following rules: tail(cons(h, t)) -> t map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (164) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (165) Obligation: Q DP problem: The TRS P consists of the following rules: IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, x1, sndsplit(y5, app(map_f(three, x0), y4)), y5) IF_8(y0, y1, y2, cons(x0, x1), y4, 0, false) -> RING(y0, y1, y2, x1, app(map_f(three, x0), y4), 0) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) tail(cons(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (166) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. tail(cons(x0, x1)) ---------------------------------------- (167) Obligation: Q DP problem: The TRS P consists of the following rules: IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, x1, sndsplit(y5, app(map_f(three, x0), y4)), y5) IF_8(y0, y1, y2, cons(x0, x1), y4, 0, false) -> RING(y0, y1, y2, x1, app(map_f(three, x0), y4), 0) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (168) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule IF_6(y0, y1, y2, cons(x0, x1), y4, y5, false) -> IF_8(y0, y1, y2, cons(x0, x1), y4, y5, empty(fstsplit(y5, app(map_f(three, x0), y4)))) we obtained the following new rules [LPAR04]: (IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))),IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil))))) (IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))),IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5)))))) ---------------------------------------- (169) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, x1, sndsplit(y5, app(map_f(three, x0), y4)), y5) IF_8(y0, y1, y2, cons(x0, x1), y4, 0, false) -> RING(y0, y1, y2, x1, app(map_f(three, x0), y4), 0) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (170) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule IF_8(y0, y1, y2, cons(x0, x1), y4, y5, false) -> RING(y0, y1, y2, x1, sndsplit(y5, app(map_f(three, x0), y4)), y5) we obtained the following new rules [LPAR04]: (IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5),IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5)) (IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7),IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7)) ---------------------------------------- (171) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_8(y0, y1, y2, cons(x0, x1), y4, 0, false) -> RING(y0, y1, y2, x1, app(map_f(three, x0), y4), 0) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (172) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule IF_8(y0, y1, y2, cons(x0, x1), y4, 0, false) -> RING(y0, y1, y2, x1, app(map_f(three, x0), y4), 0) we obtained the following new rules [LPAR04]: (IF_8(z0, z1, z2, cons(z3, z4), nil, 0, false) -> RING(z0, z1, z2, z4, app(map_f(three, z3), nil), 0),IF_8(z0, z1, z2, cons(z3, z4), nil, 0, false) -> RING(z0, z1, z2, z4, app(map_f(three, z3), nil), 0)) (IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), 0, false) -> RING(z0, z1, z2, z4, app(map_f(three, z3), cons(z5, z6)), 0),IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), 0, false) -> RING(z0, z1, z2, z4, app(map_f(three, z3), cons(z5, z6)), 0)) ---------------------------------------- (173) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) IF_8(z0, z1, z2, cons(z3, z4), nil, 0, false) -> RING(z0, z1, z2, z4, app(map_f(three, z3), nil), 0) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), 0, false) -> RING(z0, z1, z2, z4, app(map_f(three, z3), cons(z5, z6)), 0) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (174) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule RING(y0, y1, y2, y3, nil, y5) -> IF_6(y0, y1, y2, y3, nil, y5, leq(y5, 0)) we obtained the following new rules [LPAR04]: (RING(x0, x1, x2, cons(y_3, y_4), nil, x4) -> IF_6(x0, x1, x2, cons(y_3, y_4), nil, x4, leq(x4, 0)),RING(x0, x1, x2, cons(y_3, y_4), nil, x4) -> IF_6(x0, x1, x2, cons(y_3, y_4), nil, x4, leq(x4, 0))) ---------------------------------------- (175) Obligation: Q DP problem: The TRS P consists of the following rules: RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) IF_8(z0, z1, z2, cons(z3, z4), nil, 0, false) -> RING(z0, z1, z2, z4, app(map_f(three, z3), nil), 0) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), 0, false) -> RING(z0, z1, z2, z4, app(map_f(three, z3), cons(z5, z6)), 0) RING(x0, x1, x2, cons(y_3, y_4), nil, x4) -> IF_6(x0, x1, x2, cons(y_3, y_4), nil, x4, leq(x4, 0)) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (176) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule RING(y0, y1, y2, y3, cons(x0, x1), y5) -> IF_6(y0, y1, y2, y3, cons(x0, x1), y5, leq(y5, s(length(x1)))) we obtained the following new rules [LPAR04]: (RING(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6) -> IF_6(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6, leq(x6, s(length(x5)))),RING(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6) -> IF_6(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6, leq(x6, s(length(x5))))) ---------------------------------------- (177) Obligation: Q DP problem: The TRS P consists of the following rules: IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) IF_8(z0, z1, z2, cons(z3, z4), nil, 0, false) -> RING(z0, z1, z2, z4, app(map_f(three, z3), nil), 0) IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), 0, false) -> RING(z0, z1, z2, z4, app(map_f(three, z3), cons(z5, z6)), 0) RING(x0, x1, x2, cons(y_3, y_4), nil, x4) -> IF_6(x0, x1, x2, cons(y_3, y_4), nil, x4, leq(x4, 0)) RING(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6) -> IF_6(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6, leq(x6, s(length(x5)))) The TRS R consists of the following rules: map_f(pid, nil) -> nil map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t)) app(nil, x) -> x app(cons(h, t), x) -> cons(h, app(t, x)) sndsplit(0, x) -> x sndsplit(s(n), nil) -> nil length(nil) -> 0 length(cons(h, t)) -> s(length(t)) leq(0, m) -> true leq(s(n), s(m)) -> leq(n, m) leq(s(n), 0) -> false fstsplit(0, x) -> nil fstsplit(s(n), nil) -> nil fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t)) empty(nil) -> true empty(cons(h, t)) -> false The set Q consists of the following terms: fstsplit(0, x0) fstsplit(s(x0), nil) fstsplit(s(x0), cons(x1, x2)) sndsplit(0, x0) sndsplit(s(x0), nil) sndsplit(s(x0), cons(x1, x2)) empty(nil) empty(cons(x0, x1)) leq(0, x0) leq(s(x0), 0) leq(s(x0), s(x1)) length(nil) length(cons(x0, x1)) app(nil, x0) app(cons(x0, x1), x2) map_f(x0, nil) map_f(x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (178) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *RING(x0, x1, x2, cons(y_3, y_4), nil, x4) -> IF_6(x0, x1, x2, cons(y_3, y_4), nil, x4, leq(x4, 0)) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *RING(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6) -> IF_6(x0, x1, x2, cons(y_3, y_4), cons(x4, x5), x6, leq(x6, s(length(x5)))) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *IF_6(z0, z1, z2, cons(x3, x4), nil, z4, false) -> IF_8(z0, z1, z2, cons(x3, x4), nil, z4, empty(fstsplit(z4, app(map_f(three, x3), nil)))) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *IF_6(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, false) -> IF_8(z0, z1, z2, cons(x3, x4), cons(z4, z5), z6, empty(fstsplit(z6, app(map_f(three, x3), cons(z4, z5))))) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4, 5 >= 5, 6 >= 6 *IF_8(z0, z1, z2, cons(z3, z4), nil, z5, false) -> RING(z0, z1, z2, z4, sndsplit(z5, app(map_f(three, z3), nil)), z5) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 > 4, 6 >= 6 *IF_8(z0, z1, z2, cons(z3, z4), nil, 0, false) -> RING(z0, z1, z2, z4, app(map_f(three, z3), nil), 0) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 > 4, 6 >= 6 *IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), z7, false) -> RING(z0, z1, z2, z4, sndsplit(z7, app(map_f(three, z3), cons(z5, z6))), z7) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 > 4, 6 >= 6 *IF_8(z0, z1, z2, cons(z3, z4), cons(z5, z6), 0, false) -> RING(z0, z1, z2, z4, app(map_f(three, z3), cons(z5, z6)), 0) The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 > 4, 6 >= 6 ---------------------------------------- (179) YES