/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Quasi decreasingness of the given CTRS could be proven: (0) CTRS (1) CTRSToQTRSProof [SOUND, 0 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 48 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 0 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (10) QDP (11) TransformationProof [EQUIVALENT, 0 ms] (12) QDP (13) TransformationProof [EQUIVALENT, 0 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) QDP (17) TransformationProof [EQUIVALENT, 0 ms] (18) QDP (19) DependencyGraphProof [EQUIVALENT, 0 ms] (20) QDP (21) TransformationProof [EQUIVALENT, 0 ms] (22) QDP (23) DependencyGraphProof [EQUIVALENT, 0 ms] (24) QDP (25) TransformationProof [EQUIVALENT, 0 ms] (26) QDP (27) DependencyGraphProof [EQUIVALENT, 0 ms] (28) QDP (29) TransformationProof [EQUIVALENT, 0 ms] (30) QDP (31) DependencyGraphProof [EQUIVALENT, 0 ms] (32) QDP (33) TransformationProof [EQUIVALENT, 0 ms] (34) QDP (35) DependencyGraphProof [EQUIVALENT, 0 ms] (36) QDP (37) TransformationProof [EQUIVALENT, 0 ms] (38) QDP (39) DependencyGraphProof [EQUIVALENT, 0 ms] (40) QDP (41) TransformationProof [EQUIVALENT, 0 ms] (42) QDP (43) DependencyGraphProof [EQUIVALENT, 0 ms] (44) QDP (45) TransformationProof [EQUIVALENT, 0 ms] (46) QDP (47) DependencyGraphProof [EQUIVALENT, 0 ms] (48) QDP (49) TransformationProof [EQUIVALENT, 0 ms] (50) QDP (51) TransformationProof [EQUIVALENT, 0 ms] (52) QDP (53) DependencyGraphProof [EQUIVALENT, 0 ms] (54) QDP (55) TransformationProof [EQUIVALENT, 0 ms] (56) QDP (57) DependencyGraphProof [EQUIVALENT, 0 ms] (58) QDP (59) TransformationProof [EQUIVALENT, 0 ms] (60) QDP (61) DependencyGraphProof [EQUIVALENT, 0 ms] (62) QDP (63) TransformationProof [EQUIVALENT, 0 ms] (64) QDP (65) DependencyGraphProof [EQUIVALENT, 0 ms] (66) QDP (67) TransformationProof [EQUIVALENT, 0 ms] (68) QDP (69) TransformationProof [EQUIVALENT, 0 ms] (70) QDP (71) DependencyGraphProof [EQUIVALENT, 0 ms] (72) QDP (73) TransformationProof [EQUIVALENT, 0 ms] (74) QDP (75) DependencyGraphProof [EQUIVALENT, 0 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) DependencyGraphProof [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) DependencyGraphProof [EQUIVALENT, 0 ms] (92) QDP (93) TransformationProof [EQUIVALENT, 0 ms] (94) QDP (95) DependencyGraphProof [EQUIVALENT, 0 ms] (96) QDP (97) TransformationProof [EQUIVALENT, 0 ms] (98) QDP (99) DependencyGraphProof [EQUIVALENT, 0 ms] (100) QDP (101) TransformationProof [EQUIVALENT, 0 ms] (102) QDP (103) DependencyGraphProof [EQUIVALENT, 0 ms] (104) QDP (105) TransformationProof [EQUIVALENT, 0 ms] (106) QDP (107) DependencyGraphProof [EQUIVALENT, 0 ms] (108) QDP (109) TransformationProof [EQUIVALENT, 0 ms] (110) QDP (111) DependencyGraphProof [EQUIVALENT, 0 ms] (112) QDP (113) TransformationProof [EQUIVALENT, 0 ms] (114) QDP (115) DependencyGraphProof [EQUIVALENT, 0 ms] (116) QDP (117) TransformationProof [EQUIVALENT, 0 ms] (118) QDP (119) DependencyGraphProof [EQUIVALENT, 0 ms] (120) QDP (121) TransformationProof [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) DependencyGraphProof [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) TransformationProof [EQUIVALENT, 0 ms] (138) QDP (139) DependencyGraphProof [EQUIVALENT, 0 ms] (140) QDP (141) TransformationProof [EQUIVALENT, 0 ms] (142) QDP (143) DependencyGraphProof [EQUIVALENT, 0 ms] (144) QDP (145) TransformationProof [EQUIVALENT, 0 ms] (146) QDP (147) DependencyGraphProof [EQUIVALENT, 0 ms] (148) QDP (149) TransformationProof [EQUIVALENT, 0 ms] (150) QDP (151) DependencyGraphProof [EQUIVALENT, 0 ms] (152) QDP (153) TransformationProof [EQUIVALENT, 0 ms] (154) QDP (155) DependencyGraphProof [EQUIVALENT, 0 ms] (156) QDP (157) TransformationProof [EQUIVALENT, 0 ms] (158) QDP (159) DependencyGraphProof [EQUIVALENT, 0 ms] (160) QDP (161) TransformationProof [EQUIVALENT, 0 ms] (162) QDP (163) DependencyGraphProof [EQUIVALENT, 0 ms] (164) QDP (165) TransformationProof [EQUIVALENT, 0 ms] (166) QDP (167) DependencyGraphProof [EQUIVALENT, 0 ms] (168) QDP (169) TransformationProof [EQUIVALENT, 0 ms] (170) QDP (171) TransformationProof [EQUIVALENT, 0 ms] (172) QDP (173) TransformationProof [EQUIVALENT, 0 ms] (174) QDP (175) TransformationProof [EQUIVALENT, 0 ms] (176) QDP (177) DependencyGraphProof [EQUIVALENT, 0 ms] (178) QDP (179) TransformationProof [EQUIVALENT, 0 ms] (180) QDP (181) TransformationProof [EQUIVALENT, 0 ms] (182) QDP (183) TransformationProof [EQUIVALENT, 0 ms] (184) QDP (185) TransformationProof [EQUIVALENT, 0 ms] (186) QDP (187) TransformationProof [EQUIVALENT, 0 ms] (188) QDP (189) DependencyGraphProof [EQUIVALENT, 0 ms] (190) QDP (191) TransformationProof [EQUIVALENT, 0 ms] (192) QDP (193) DependencyGraphProof [EQUIVALENT, 0 ms] (194) QDP (195) TransformationProof [EQUIVALENT, 0 ms] (196) QDP (197) DependencyGraphProof [EQUIVALENT, 0 ms] (198) QDP (199) TransformationProof [EQUIVALENT, 0 ms] (200) QDP (201) DependencyGraphProof [EQUIVALENT, 0 ms] (202) QDP (203) TransformationProof [EQUIVALENT, 0 ms] (204) QDP (205) DependencyGraphProof [EQUIVALENT, 0 ms] (206) QDP (207) TransformationProof [EQUIVALENT, 0 ms] (208) QDP (209) DependencyGraphProof [EQUIVALENT, 0 ms] (210) QDP (211) TransformationProof [EQUIVALENT, 0 ms] (212) QDP (213) TransformationProof [EQUIVALENT, 0 ms] (214) QDP (215) DependencyGraphProof [EQUIVALENT, 0 ms] (216) QDP (217) TransformationProof [EQUIVALENT, 0 ms] (218) QDP (219) TransformationProof [EQUIVALENT, 0 ms] (220) QDP (221) TransformationProof [EQUIVALENT, 0 ms] (222) QDP (223) DependencyGraphProof [EQUIVALENT, 0 ms] (224) QDP (225) TransformationProof [EQUIVALENT, 0 ms] (226) QDP (227) TransformationProof [EQUIVALENT, 0 ms] (228) QDP (229) TransformationProof [EQUIVALENT, 0 ms] (230) QDP (231) TransformationProof [EQUIVALENT, 0 ms] (232) QDP (233) TransformationProof [EQUIVALENT, 0 ms] (234) QDP (235) DependencyGraphProof [EQUIVALENT, 0 ms] (236) QDP (237) TransformationProof [EQUIVALENT, 0 ms] (238) QDP (239) DependencyGraphProof [EQUIVALENT, 0 ms] (240) QDP (241) TransformationProof [EQUIVALENT, 0 ms] (242) QDP (243) DependencyGraphProof [EQUIVALENT, 0 ms] (244) QDP (245) TransformationProof [EQUIVALENT, 0 ms] (246) QDP (247) TransformationProof [EQUIVALENT, 0 ms] (248) QDP (249) DependencyGraphProof [EQUIVALENT, 0 ms] (250) QDP (251) TransformationProof [EQUIVALENT, 0 ms] (252) QDP (253) TransformationProof [EQUIVALENT, 0 ms] (254) QDP (255) TransformationProof [EQUIVALENT, 0 ms] (256) QDP (257) DependencyGraphProof [EQUIVALENT, 0 ms] (258) QDP (259) UsableRulesProof [EQUIVALENT, 0 ms] (260) QDP (261) MNOCProof [EQUIVALENT, 2 ms] (262) QDP (263) DependencyGraphProof [EQUIVALENT, 0 ms] (264) QDP (265) UsableRulesProof [EQUIVALENT, 0 ms] (266) QDP (267) QReductionProof [EQUIVALENT, 0 ms] (268) QDP (269) TransformationProof [EQUIVALENT, 0 ms] (270) QDP (271) DependencyGraphProof [EQUIVALENT, 0 ms] (272) TRUE ---------------------------------------- (0) Obligation: Conditional term rewrite system: The TRS R consists of the following rules: or(0, x) -> x or(x, 0) -> x or(1, x) -> 1 or(x, 1) -> 1 or(x, not(x)) -> 1 or(not(x), x) -> 1 and(0, x) -> 0 and(x, 0) -> 0 and(1, x) -> x and(x, 1) -> x and(x, not(x)) -> 0 and(not(x), x) -> 0 not(1) -> 0 not(0) -> 1 The conditional TRS C consists of the following conditional rules: implies(x, y) -> 1 <= not(x) -> 1 implies(x, y) -> 1 <= y -> 1 implies(x, y) -> 0 <= x -> 1, y -> 0 f(x) -> f(0) <= implies(implies(x, implies(x, 0)), 0) -> 1 ---------------------------------------- (1) CTRSToQTRSProof (SOUND) The conditional rules have been transormed into unconditional rules according to [CTRS,AAECCNOC]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) U1(1) -> 1 implies(x, y) -> U2(y) U2(1) -> 1 implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 f(x) -> U5(implies(implies(x, implies(x, 0)), 0)) U5(1) -> f(0) or(0, x) -> x or(x, 0) -> x or(1, x) -> 1 or(x, 1) -> 1 or(x, not(x)) -> 1 or(not(x), x) -> 1 and(0, x) -> 0 and(x, 0) -> 0 and(1, x) -> x and(x, 1) -> x and(x, not(x)) -> 0 and(not(x), x) -> 0 not(1) -> 0 not(0) -> 1 Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(1) = 0 POL(U1(x_1)) = x_1 POL(U2(x_1)) = x_1 POL(U3(x_1, x_2)) = x_1 + x_2 POL(U4(x_1)) = x_1 POL(U5(x_1)) = x_1 POL(and(x_1, x_2)) = 2 + x_1 + 2*x_2 POL(f(x_1)) = 2*x_1 POL(implies(x_1, x_2)) = x_1 + x_2 POL(not(x_1)) = x_1 POL(or(x_1, x_2)) = 1 + x_1 + x_2 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: or(0, x) -> x or(x, 0) -> x or(1, x) -> 1 or(x, 1) -> 1 or(x, not(x)) -> 1 or(not(x), x) -> 1 and(0, x) -> 0 and(x, 0) -> 0 and(1, x) -> x and(x, 1) -> x and(x, not(x)) -> 0 and(not(x), x) -> 0 ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) U1(1) -> 1 implies(x, y) -> U2(y) U2(1) -> 1 implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 f(x) -> U5(implies(implies(x, implies(x, 0)), 0)) U5(1) -> f(0) not(1) -> 0 not(0) -> 1 Q is empty. ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: IMPLIES(x, y) -> U1^1(not(x)) IMPLIES(x, y) -> NOT(x) IMPLIES(x, y) -> U2^1(y) IMPLIES(x, y) -> U3^1(x, y) U3^1(1, y) -> U4^1(y) F(x) -> U5^1(implies(implies(x, implies(x, 0)), 0)) F(x) -> IMPLIES(implies(x, implies(x, 0)), 0) F(x) -> IMPLIES(x, implies(x, 0)) F(x) -> IMPLIES(x, 0) U5^1(1) -> F(0) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) U1(1) -> 1 implies(x, y) -> U2(y) U2(1) -> 1 implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 f(x) -> U5(implies(implies(x, implies(x, 0)), 0)) U5(1) -> f(0) not(1) -> 0 not(0) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 8 less nodes. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: F(x) -> U5^1(implies(implies(x, implies(x, 0)), 0)) U5^1(1) -> F(0) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) U1(1) -> 1 implies(x, y) -> U2(y) U2(1) -> 1 implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 f(x) -> U5(implies(implies(x, implies(x, 0)), 0)) U5(1) -> f(0) not(1) -> 0 not(0) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) UsableRulesReductionPairsProof (EQUIVALENT) By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well. No dependency pairs are removed. The following rules are removed from R: f(x) -> U5(implies(implies(x, implies(x, 0)), 0)) U5(1) -> f(0) Used ordering: POLO with Polynomial interpretation [POLO]: POL(0) = 0 POL(1) = 0 POL(F(x_1)) = 2*x_1 POL(U1(x_1)) = x_1 POL(U2(x_1)) = x_1 POL(U3(x_1, x_2)) = x_1 + x_2 POL(U4(x_1)) = x_1 POL(U5^1(x_1)) = x_1 POL(implies(x_1, x_2)) = x_1 + x_2 POL(not(x_1)) = x_1 ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: F(x) -> U5^1(implies(implies(x, implies(x, 0)), 0)) U5^1(1) -> F(0) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule F(x) -> U5^1(implies(implies(x, implies(x, 0)), 0)) we obtained the following new rules [LPAR04]: (F(0) -> U5^1(implies(implies(0, implies(0, 0)), 0)),F(0) -> U5^1(implies(implies(0, implies(0, 0)), 0))) ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(implies(0, implies(0, 0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(implies(0, implies(0, 0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(implies(0, implies(0, 0))))),F(0) -> U5^1(U1(not(implies(0, implies(0, 0)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(implies(0, implies(0, 0)), 0)),F(0) -> U5^1(U3(implies(0, implies(0, 0)), 0))) (F(0) -> U5^1(implies(U1(not(0)), 0)),F(0) -> U5^1(implies(U1(not(0)), 0))) (F(0) -> U5^1(implies(U2(implies(0, 0)), 0)),F(0) -> U5^1(implies(U2(implies(0, 0)), 0))) (F(0) -> U5^1(implies(U3(0, implies(0, 0)), 0)),F(0) -> U5^1(implies(U3(0, implies(0, 0)), 0))) (F(0) -> U5^1(implies(implies(0, U1(not(0))), 0)),F(0) -> U5^1(implies(implies(0, U1(not(0))), 0))) (F(0) -> U5^1(implies(implies(0, U2(0)), 0)),F(0) -> U5^1(implies(implies(0, U2(0)), 0))) (F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)),F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0))) ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(implies(0, implies(0, 0))))) F(0) -> U5^1(U2(0)) F(0) -> U5^1(U3(implies(0, implies(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(U2(implies(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, implies(0, 0)), 0)) F(0) -> U5^1(implies(implies(0, U1(not(0))), 0)) F(0) -> U5^1(implies(implies(0, U2(0)), 0)) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(implies(0, implies(0, 0))))) U5^1(1) -> F(0) F(0) -> U5^1(U3(implies(0, implies(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(U2(implies(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, implies(0, 0)), 0)) F(0) -> U5^1(implies(implies(0, U1(not(0))), 0)) F(0) -> U5^1(implies(implies(0, U2(0)), 0)) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(implies(0, implies(0, 0))))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(not(0))))),F(0) -> U5^1(U1(not(U1(not(0)))))) (F(0) -> U5^1(U1(not(U2(implies(0, 0))))),F(0) -> U5^1(U1(not(U2(implies(0, 0)))))) (F(0) -> U5^1(U1(not(U3(0, implies(0, 0))))),F(0) -> U5^1(U1(not(U3(0, implies(0, 0)))))) (F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))),F(0) -> U5^1(U1(not(implies(0, U1(not(0))))))) (F(0) -> U5^1(U1(not(implies(0, U2(0))))),F(0) -> U5^1(U1(not(implies(0, U2(0)))))) (F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))),F(0) -> U5^1(U1(not(implies(0, U3(0, 0)))))) ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(implies(0, implies(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(U2(implies(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, implies(0, 0)), 0)) F(0) -> U5^1(implies(implies(0, U1(not(0))), 0)) F(0) -> U5^1(implies(implies(0, U2(0)), 0)) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(U3(0, implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U3(implies(0, implies(0, 0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(U2(implies(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, implies(0, 0)), 0)) F(0) -> U5^1(implies(implies(0, U1(not(0))), 0)) F(0) -> U5^1(implies(implies(0, U2(0)), 0)) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(implies(0, implies(0, 0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(U1(not(0)), 0)),F(0) -> U5^1(U3(U1(not(0)), 0))) (F(0) -> U5^1(U3(U2(implies(0, 0)), 0)),F(0) -> U5^1(U3(U2(implies(0, 0)), 0))) (F(0) -> U5^1(U3(U3(0, implies(0, 0)), 0)),F(0) -> U5^1(U3(U3(0, implies(0, 0)), 0))) (F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)),F(0) -> U5^1(U3(implies(0, U1(not(0))), 0))) (F(0) -> U5^1(U3(implies(0, U2(0)), 0)),F(0) -> U5^1(U3(implies(0, U2(0)), 0))) (F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)),F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0))) ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(U2(implies(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, implies(0, 0)), 0)) F(0) -> U5^1(implies(implies(0, U1(not(0))), 0)) F(0) -> U5^1(implies(implies(0, U2(0)), 0)) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(U3(0, implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U1(not(0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U2(implies(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, implies(0, 0)), 0)) F(0) -> U5^1(implies(implies(0, U1(not(0))), 0)) F(0) -> U5^1(implies(implies(0, U2(0)), 0)) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U1(not(0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(not(0))))),F(0) -> U5^1(U1(not(U1(not(0)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U1(not(0)), 0)),F(0) -> U5^1(U3(U1(not(0)), 0))) (F(0) -> U5^1(implies(U1(1), 0)),F(0) -> U5^1(implies(U1(1), 0))) ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U2(implies(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, implies(0, 0)), 0)) F(0) -> U5^1(implies(implies(0, U1(not(0))), 0)) F(0) -> U5^1(implies(implies(0, U2(0)), 0)) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(U2(0)) F(0) -> U5^1(implies(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U2(implies(0, 0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U3(0, implies(0, 0)), 0)) F(0) -> U5^1(implies(implies(0, U1(not(0))), 0)) F(0) -> U5^1(implies(implies(0, U2(0)), 0)) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U2(implies(0, 0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U2(implies(0, 0))))),F(0) -> U5^1(U1(not(U2(implies(0, 0)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U2(implies(0, 0)), 0)),F(0) -> U5^1(U3(U2(implies(0, 0)), 0))) (F(0) -> U5^1(implies(U2(U1(not(0))), 0)),F(0) -> U5^1(implies(U2(U1(not(0))), 0))) (F(0) -> U5^1(implies(U2(U2(0)), 0)),F(0) -> U5^1(implies(U2(U2(0)), 0))) (F(0) -> U5^1(implies(U2(U3(0, 0)), 0)),F(0) -> U5^1(implies(U2(U3(0, 0)), 0))) ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U3(0, implies(0, 0)), 0)) F(0) -> U5^1(implies(implies(0, U1(not(0))), 0)) F(0) -> U5^1(implies(implies(0, U2(0)), 0)) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(U2(0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U3(0, implies(0, 0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(implies(0, U1(not(0))), 0)) F(0) -> U5^1(implies(implies(0, U2(0)), 0)) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U3(0, implies(0, 0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U3(0, implies(0, 0))))),F(0) -> U5^1(U1(not(U3(0, implies(0, 0)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U3(0, implies(0, 0)), 0)),F(0) -> U5^1(U3(U3(0, implies(0, 0)), 0))) (F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)),F(0) -> U5^1(implies(U3(0, U1(not(0))), 0))) (F(0) -> U5^1(implies(U3(0, U2(0)), 0)),F(0) -> U5^1(implies(U3(0, U2(0)), 0))) (F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)),F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0))) ---------------------------------------- (34) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(implies(0, U1(not(0))), 0)) F(0) -> U5^1(implies(implies(0, U2(0)), 0)) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(U1(not(U3(0, implies(0, 0))))) F(0) -> U5^1(U2(0)) F(0) -> U5^1(U3(U3(0, implies(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (35) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (36) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(implies(0, U1(not(0))), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(implies(0, U2(0)), 0)) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (37) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(implies(0, U1(not(0))), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))),F(0) -> U5^1(U1(not(implies(0, U1(not(0))))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)),F(0) -> U5^1(U3(implies(0, U1(not(0))), 0))) (F(0) -> U5^1(implies(U1(not(0)), 0)),F(0) -> U5^1(implies(U1(not(0)), 0))) (F(0) -> U5^1(implies(U2(U1(not(0))), 0)),F(0) -> U5^1(implies(U2(U1(not(0))), 0))) (F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)),F(0) -> U5^1(implies(U3(0, U1(not(0))), 0))) (F(0) -> U5^1(implies(implies(0, U1(1)), 0)),F(0) -> U5^1(implies(implies(0, U1(1)), 0))) ---------------------------------------- (38) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(implies(0, U2(0)), 0)) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(U2(0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (39) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (40) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(implies(0, U2(0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (41) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(implies(0, U2(0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(implies(0, U2(0))))),F(0) -> U5^1(U1(not(implies(0, U2(0)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(implies(0, U2(0)), 0)),F(0) -> U5^1(U3(implies(0, U2(0)), 0))) (F(0) -> U5^1(implies(U1(not(0)), 0)),F(0) -> U5^1(implies(U1(not(0)), 0))) (F(0) -> U5^1(implies(U2(U2(0)), 0)),F(0) -> U5^1(implies(U2(U2(0)), 0))) (F(0) -> U5^1(implies(U3(0, U2(0)), 0)),F(0) -> U5^1(implies(U3(0, U2(0)), 0))) ---------------------------------------- (42) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U2(0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (43) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (44) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (45) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(implies(0, U3(0, 0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))),F(0) -> U5^1(U1(not(implies(0, U3(0, 0)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)),F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0))) (F(0) -> U5^1(implies(U1(not(0)), 0)),F(0) -> U5^1(implies(U1(not(0)), 0))) (F(0) -> U5^1(implies(U2(U3(0, 0)), 0)),F(0) -> U5^1(implies(U2(U3(0, 0)), 0))) (F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)),F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0))) ---------------------------------------- (46) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U2(0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (47) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (48) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(U1(not(0))))) U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (49) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(U1(not(0))))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(1)))),F(0) -> U5^1(U1(not(U1(1))))) ---------------------------------------- (50) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U2(implies(0, 0))))) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (51) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(U2(implies(0, 0))))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U2(U1(not(0)))))),F(0) -> U5^1(U1(not(U2(U1(not(0))))))) (F(0) -> U5^1(U1(not(U2(U2(0))))),F(0) -> U5^1(U1(not(U2(U2(0)))))) (F(0) -> U5^1(U1(not(U2(U3(0, 0))))),F(0) -> U5^1(U1(not(U2(U3(0, 0)))))) ---------------------------------------- (52) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U2(U2(0))))) F(0) -> U5^1(U1(not(U2(U3(0, 0))))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (53) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (54) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) U5^1(1) -> F(0) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (55) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(implies(0, U1(not(0)))))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(not(0))))),F(0) -> U5^1(U1(not(U1(not(0)))))) (F(0) -> U5^1(U1(not(U2(U1(not(0)))))),F(0) -> U5^1(U1(not(U2(U1(not(0))))))) (F(0) -> U5^1(U1(not(U3(0, U1(not(0)))))),F(0) -> U5^1(U1(not(U3(0, U1(not(0))))))) (F(0) -> U5^1(U1(not(implies(0, U1(1))))),F(0) -> U5^1(U1(not(implies(0, U1(1)))))) ---------------------------------------- (56) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(implies(0, U2(0))))) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U3(0, U1(not(0)))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (57) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (58) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(implies(0, U2(0))))) U5^1(1) -> F(0) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (59) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(implies(0, U2(0))))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(not(0))))),F(0) -> U5^1(U1(not(U1(not(0)))))) (F(0) -> U5^1(U1(not(U2(U2(0))))),F(0) -> U5^1(U1(not(U2(U2(0)))))) (F(0) -> U5^1(U1(not(U3(0, U2(0))))),F(0) -> U5^1(U1(not(U3(0, U2(0)))))) ---------------------------------------- (60) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U1(not(U2(U2(0))))) F(0) -> U5^1(U1(not(U3(0, U2(0))))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (61) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (62) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) U5^1(1) -> F(0) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (63) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(implies(0, U3(0, 0))))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(not(0))))),F(0) -> U5^1(U1(not(U1(not(0)))))) (F(0) -> U5^1(U1(not(U2(U3(0, 0))))),F(0) -> U5^1(U1(not(U2(U3(0, 0)))))) (F(0) -> U5^1(U1(not(U3(0, U3(0, 0))))),F(0) -> U5^1(U1(not(U3(0, U3(0, 0)))))) ---------------------------------------- (64) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U1(not(U2(U3(0, 0))))) F(0) -> U5^1(U1(not(U3(0, U3(0, 0))))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (65) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (66) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U3(U1(not(0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (67) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(U1(not(0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(U1(1), 0)),F(0) -> U5^1(U3(U1(1), 0))) ---------------------------------------- (68) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (69) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(U2(implies(0, 0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(U2(U1(not(0))), 0)),F(0) -> U5^1(U3(U2(U1(not(0))), 0))) (F(0) -> U5^1(U3(U2(U2(0)), 0)),F(0) -> U5^1(U3(U2(U2(0)), 0))) (F(0) -> U5^1(U3(U2(U3(0, 0)), 0)),F(0) -> U5^1(U3(U2(U3(0, 0)), 0))) ---------------------------------------- (70) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U2(U2(0)), 0)) F(0) -> U5^1(U3(U2(U3(0, 0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (71) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (72) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) U5^1(1) -> F(0) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (73) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(implies(0, U1(not(0))), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(U1(not(0)), 0)),F(0) -> U5^1(U3(U1(not(0)), 0))) (F(0) -> U5^1(U3(U2(U1(not(0))), 0)),F(0) -> U5^1(U3(U2(U1(not(0))), 0))) (F(0) -> U5^1(U3(U3(0, U1(not(0))), 0)),F(0) -> U5^1(U3(U3(0, U1(not(0))), 0))) (F(0) -> U5^1(U3(implies(0, U1(1)), 0)),F(0) -> U5^1(U3(implies(0, U1(1)), 0))) ---------------------------------------- (74) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(implies(0, U2(0)), 0)) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U3(0, U1(not(0))), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (75) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (76) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U3(implies(0, U2(0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (77) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(implies(0, U2(0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(U1(not(0)), 0)),F(0) -> U5^1(U3(U1(not(0)), 0))) (F(0) -> U5^1(U3(U2(U2(0)), 0)),F(0) -> U5^1(U3(U2(U2(0)), 0))) (F(0) -> U5^1(U3(U3(0, U2(0)), 0)),F(0) -> U5^1(U3(U3(0, U2(0)), 0))) ---------------------------------------- (78) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(U3(U2(U2(0)), 0)) F(0) -> U5^1(U3(U3(0, U2(0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. 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 2 less nodes. ---------------------------------------- (80) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (81) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(implies(0, U3(0, 0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(U1(not(0)), 0)),F(0) -> U5^1(U3(U1(not(0)), 0))) (F(0) -> U5^1(U3(U2(U3(0, 0)), 0)),F(0) -> U5^1(U3(U2(U3(0, 0)), 0))) (F(0) -> U5^1(U3(U3(0, U3(0, 0)), 0)),F(0) -> U5^1(U3(U3(0, U3(0, 0)), 0))) ---------------------------------------- (82) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(U3(U2(U3(0, 0)), 0)) F(0) -> U5^1(U3(U3(0, U3(0, 0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (83) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (84) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U1(1), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (85) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U1(1), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(1)))),F(0) -> U5^1(U1(not(U1(1))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U1(1), 0)),F(0) -> U5^1(U3(U1(1), 0))) (F(0) -> U5^1(implies(1, 0)),F(0) -> U5^1(implies(1, 0))) ---------------------------------------- (86) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U2(U1(not(0))), 0)) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(U2(0)) F(0) -> U5^1(implies(1, 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. 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: F(0) -> U5^1(implies(U2(U1(not(0))), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (89) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U2(U1(not(0))), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U2(U1(not(0)))))),F(0) -> U5^1(U1(not(U2(U1(not(0))))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U2(U1(not(0))), 0)),F(0) -> U5^1(U3(U2(U1(not(0))), 0))) (F(0) -> U5^1(implies(U2(U1(1)), 0)),F(0) -> U5^1(implies(U2(U1(1)), 0))) ---------------------------------------- (90) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U2(U2(0)), 0)) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(U2(0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (91) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (92) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U2(U2(0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (93) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U2(U2(0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U2(U2(0))))),F(0) -> U5^1(U1(not(U2(U2(0)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U2(U2(0)), 0)),F(0) -> U5^1(U3(U2(U2(0)), 0))) ---------------------------------------- (94) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(U1(not(U2(U2(0))))) F(0) -> U5^1(U2(0)) F(0) -> U5^1(U3(U2(U2(0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (95) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (96) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (97) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U2(U3(0, 0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U2(U3(0, 0))))),F(0) -> U5^1(U1(not(U2(U3(0, 0)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U2(U3(0, 0)), 0)),F(0) -> U5^1(U3(U2(U3(0, 0)), 0))) ---------------------------------------- (98) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(U1(not(U2(U3(0, 0))))) F(0) -> U5^1(U2(0)) F(0) -> U5^1(U3(U2(U3(0, 0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (99) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (100) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (101) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U3(0, U1(not(0))), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U3(0, U1(not(0)))))),F(0) -> U5^1(U1(not(U3(0, U1(not(0))))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U3(0, U1(not(0))), 0)),F(0) -> U5^1(U3(U3(0, U1(not(0))), 0))) (F(0) -> U5^1(implies(U3(0, U1(1)), 0)),F(0) -> U5^1(implies(U3(0, U1(1)), 0))) ---------------------------------------- (102) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U3(0, U2(0)), 0)) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(U1(not(U3(0, U1(not(0)))))) F(0) -> U5^1(U2(0)) F(0) -> U5^1(U3(U3(0, U1(not(0))), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (103) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (104) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U3(0, U2(0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (105) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U3(0, U2(0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U3(0, U2(0))))),F(0) -> U5^1(U1(not(U3(0, U2(0)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U3(0, U2(0)), 0)),F(0) -> U5^1(U3(U3(0, U2(0)), 0))) ---------------------------------------- (106) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U3(0, U2(0))))) F(0) -> U5^1(U2(0)) F(0) -> U5^1(U3(U3(0, U2(0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (107) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (108) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (109) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U3(0, U3(0, 0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U3(0, U3(0, 0))))),F(0) -> U5^1(U1(not(U3(0, U3(0, 0)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U3(0, U3(0, 0)), 0)),F(0) -> U5^1(U3(U3(0, U3(0, 0)), 0))) ---------------------------------------- (110) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U3(0, U3(0, 0))))) F(0) -> U5^1(U2(0)) F(0) -> U5^1(U3(U3(0, U3(0, 0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (111) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (112) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U1(not(0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (113) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U1(not(0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(not(0))))),F(0) -> U5^1(U1(not(U1(not(0)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U1(not(0)), 0)),F(0) -> U5^1(U3(U1(not(0)), 0))) (F(0) -> U5^1(implies(U1(1), 0)),F(0) -> U5^1(implies(U1(1), 0))) ---------------------------------------- (114) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(implies(0, U1(1)), 0)) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(U2(0)) F(0) -> U5^1(implies(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (115) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (116) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(implies(0, U1(1)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (117) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(implies(0, U1(1)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(implies(0, U1(1))))),F(0) -> U5^1(U1(not(implies(0, U1(1)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(implies(0, U1(1)), 0)),F(0) -> U5^1(U3(implies(0, U1(1)), 0))) (F(0) -> U5^1(implies(U1(not(0)), 0)),F(0) -> U5^1(implies(U1(not(0)), 0))) (F(0) -> U5^1(implies(U2(U1(1)), 0)),F(0) -> U5^1(implies(U2(U1(1)), 0))) (F(0) -> U5^1(implies(U3(0, U1(1)), 0)),F(0) -> U5^1(implies(U3(0, U1(1)), 0))) (F(0) -> U5^1(implies(implies(0, 1), 0)),F(0) -> U5^1(implies(implies(0, 1), 0))) ---------------------------------------- (118) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(U2(0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (119) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (120) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(U1(1)))) U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (121) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(U1(1)))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(1))),F(0) -> U5^1(U1(not(1)))) ---------------------------------------- (122) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U2(U1(not(0)))))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (123) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(U2(U1(not(0)))))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U2(U1(1))))),F(0) -> U5^1(U1(not(U2(U1(1)))))) ---------------------------------------- (124) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (125) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(U1(not(0))))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(1)))),F(0) -> U5^1(U1(not(U1(1))))) ---------------------------------------- (126) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(implies(0, U1(1))))) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (127) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(implies(0, U1(1))))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(not(0))))),F(0) -> U5^1(U1(not(U1(not(0)))))) (F(0) -> U5^1(U1(not(U2(U1(1))))),F(0) -> U5^1(U1(not(U2(U1(1)))))) (F(0) -> U5^1(U1(not(U3(0, U1(1))))),F(0) -> U5^1(U1(not(U3(0, U1(1)))))) (F(0) -> U5^1(U1(not(implies(0, 1)))),F(0) -> U5^1(U1(not(implies(0, 1))))) ---------------------------------------- (128) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U3(0, U1(1))))) F(0) -> U5^1(U1(not(implies(0, 1)))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (129) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (130) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U3(U1(1), 0)) U5^1(1) -> F(0) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (131) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(U1(1), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(1, 0)),F(0) -> U5^1(U3(1, 0))) ---------------------------------------- (132) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(U2(U1(not(0))), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (133) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(U2(U1(not(0))), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(U2(U1(1)), 0)),F(0) -> U5^1(U3(U2(U1(1)), 0))) ---------------------------------------- (134) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (135) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(U1(not(0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(U1(1), 0)),F(0) -> U5^1(U3(U1(1), 0))) ---------------------------------------- (136) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(implies(0, U1(1)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (137) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(implies(0, U1(1)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(U1(not(0)), 0)),F(0) -> U5^1(U3(U1(not(0)), 0))) (F(0) -> U5^1(U3(U2(U1(1)), 0)),F(0) -> U5^1(U3(U2(U1(1)), 0))) (F(0) -> U5^1(U3(U3(0, U1(1)), 0)),F(0) -> U5^1(U3(U3(0, U1(1)), 0))) (F(0) -> U5^1(U3(implies(0, 1), 0)),F(0) -> U5^1(U3(implies(0, 1), 0))) ---------------------------------------- (138) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U3(0, U1(1)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (139) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (140) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(1, 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (141) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(1, 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(1))),F(0) -> U5^1(U1(not(1)))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(1, 0)),F(0) -> U5^1(U3(1, 0))) ---------------------------------------- (142) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U2(U1(1)), 0)) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(U2(0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (143) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (144) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U2(U1(1)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (145) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U2(U1(1)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U2(U1(1))))),F(0) -> U5^1(U1(not(U2(U1(1)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U2(U1(1)), 0)),F(0) -> U5^1(U3(U2(U1(1)), 0))) (F(0) -> U5^1(implies(U2(1), 0)),F(0) -> U5^1(implies(U2(1), 0))) ---------------------------------------- (146) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(U2(0)) F(0) -> U5^1(implies(U2(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (147) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (148) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U3(0, U1(1)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (149) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U3(0, U1(1)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U3(0, U1(1))))),F(0) -> U5^1(U1(not(U3(0, U1(1)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U3(0, U1(1)), 0)),F(0) -> U5^1(U3(U3(0, U1(1)), 0))) (F(0) -> U5^1(implies(U3(0, 1), 0)),F(0) -> U5^1(implies(U3(0, 1), 0))) ---------------------------------------- (150) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(U1(not(U3(0, U1(1))))) F(0) -> U5^1(U2(0)) F(0) -> U5^1(U3(U3(0, U1(1)), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (151) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (152) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U1(1), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (153) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U1(1), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(1)))),F(0) -> U5^1(U1(not(U1(1))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U1(1), 0)),F(0) -> U5^1(U3(U1(1), 0))) (F(0) -> U5^1(implies(1, 0)),F(0) -> U5^1(implies(1, 0))) ---------------------------------------- (154) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(U2(0)) F(0) -> U5^1(implies(1, 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (155) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (156) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U1(not(0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (157) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U1(not(0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(not(0))))),F(0) -> U5^1(U1(not(U1(not(0)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U1(not(0)), 0)),F(0) -> U5^1(U3(U1(not(0)), 0))) (F(0) -> U5^1(implies(U1(1), 0)),F(0) -> U5^1(implies(U1(1), 0))) ---------------------------------------- (158) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(implies(0, 1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(U2(0)) F(0) -> U5^1(implies(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (159) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (160) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(implies(0, 1), 0)) U5^1(1) -> F(0) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (161) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(implies(0, 1), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(implies(0, 1)))),F(0) -> U5^1(U1(not(implies(0, 1))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(implies(0, 1), 0)),F(0) -> U5^1(U3(implies(0, 1), 0))) (F(0) -> U5^1(implies(U1(not(0)), 0)),F(0) -> U5^1(implies(U1(not(0)), 0))) (F(0) -> U5^1(implies(U2(1), 0)),F(0) -> U5^1(implies(U2(1), 0))) (F(0) -> U5^1(implies(U3(0, 1), 0)),F(0) -> U5^1(implies(U3(0, 1), 0))) ---------------------------------------- (162) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(U2(0)) F(0) -> U5^1(implies(U1(not(0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (163) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (164) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(1))) U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (165) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(1))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(0)),F(0) -> U5^1(U1(0))) ---------------------------------------- (166) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U2(U1(1))))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (167) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (168) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(U2(U1(1))))) U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (169) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(U2(U1(1))))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U2(1)))),F(0) -> U5^1(U1(not(U2(1))))) ---------------------------------------- (170) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (171) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(U1(1)))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(1))),F(0) -> U5^1(U1(not(1)))) ---------------------------------------- (172) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (173) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(U1(not(0))))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(1)))),F(0) -> U5^1(U1(not(U1(1))))) ---------------------------------------- (174) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(implies(0, 1)))) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (175) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(implies(0, 1)))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(not(0))))),F(0) -> U5^1(U1(not(U1(not(0)))))) (F(0) -> U5^1(U1(not(U2(1)))),F(0) -> U5^1(U1(not(U2(1))))) (F(0) -> U5^1(U1(not(U3(0, 1)))),F(0) -> U5^1(U1(not(U3(0, 1))))) ---------------------------------------- (176) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U1(not(U3(0, 1)))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (177) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (178) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U3(1, 0)) U5^1(1) -> F(0) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (179) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(1, 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U4(0)),F(0) -> U5^1(U4(0))) ---------------------------------------- (180) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(U2(U1(1)), 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (181) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(U2(U1(1)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(U2(1), 0)),F(0) -> U5^1(U3(U2(1), 0))) ---------------------------------------- (182) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (183) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(U1(1), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(1, 0)),F(0) -> U5^1(U3(1, 0))) ---------------------------------------- (184) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (185) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(U1(not(0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(U1(1), 0)),F(0) -> U5^1(U3(U1(1), 0))) ---------------------------------------- (186) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(implies(0, 1), 0)) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (187) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(implies(0, 1), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(U1(not(0)), 0)),F(0) -> U5^1(U3(U1(not(0)), 0))) (F(0) -> U5^1(U3(U2(1), 0)),F(0) -> U5^1(U3(U2(1), 0))) (F(0) -> U5^1(U3(U3(0, 1), 0)),F(0) -> U5^1(U3(U3(0, 1), 0))) ---------------------------------------- (188) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U2(1), 0)) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U3(U3(0, 1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (189) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (190) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U2(1), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (191) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U2(1), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U2(1)))),F(0) -> U5^1(U1(not(U2(1))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U2(1), 0)),F(0) -> U5^1(U3(U2(1), 0))) (F(0) -> U5^1(implies(1, 0)),F(0) -> U5^1(implies(1, 0))) ---------------------------------------- (192) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U3(0, 1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U2(0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (193) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (194) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U3(0, 1), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (195) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U3(0, 1), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U3(0, 1)))),F(0) -> U5^1(U1(not(U3(0, 1))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U3(0, 1), 0)),F(0) -> U5^1(U3(U3(0, 1), 0))) ---------------------------------------- (196) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U3(0, 1)))) F(0) -> U5^1(U2(0)) F(0) -> U5^1(U3(U3(0, 1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (197) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (198) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(1, 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (199) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(1, 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(1))),F(0) -> U5^1(U1(not(1)))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(1, 0)),F(0) -> U5^1(U3(1, 0))) ---------------------------------------- (200) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U2(0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (201) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (202) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U1(1), 0)) U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (203) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U1(1), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(1)))),F(0) -> U5^1(U1(not(U1(1))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U1(1), 0)),F(0) -> U5^1(U3(U1(1), 0))) (F(0) -> U5^1(implies(1, 0)),F(0) -> U5^1(implies(1, 0))) ---------------------------------------- (204) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(not(0)), 0)) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(U2(0)) F(0) -> U5^1(implies(1, 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (205) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (206) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U1(not(0)), 0)) U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(implies(1, 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (207) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U1(not(0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(not(0))))),F(0) -> U5^1(U1(not(U1(not(0)))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U1(not(0)), 0)),F(0) -> U5^1(U3(U1(not(0)), 0))) (F(0) -> U5^1(implies(U1(1), 0)),F(0) -> U5^1(implies(U1(1), 0))) ---------------------------------------- (208) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U2(1)))) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(U2(0)) F(0) -> U5^1(implies(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (209) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (210) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(U2(1)))) U5^1(1) -> F(0) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (211) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(U2(1)))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(1))),F(0) -> U5^1(U1(not(1)))) ---------------------------------------- (212) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (213) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(1))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(0)),F(0) -> U5^1(U1(0))) ---------------------------------------- (214) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(U1(0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (215) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (216) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(U1(1)))) U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (217) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(U1(1)))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(1))),F(0) -> U5^1(U1(not(1)))) ---------------------------------------- (218) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U1(not(0))))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(U1(not(1))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (219) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(U1(not(0))))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(1)))),F(0) -> U5^1(U1(not(U1(1))))) ---------------------------------------- (220) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (221) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U4(0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(0),F(0) -> U5^1(0)) ---------------------------------------- (222) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(U2(1), 0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(0) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (223) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (224) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U3(U2(1), 0)) U5^1(1) -> F(0) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (225) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(U2(1), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(1, 0)),F(0) -> U5^1(U3(1, 0))) ---------------------------------------- (226) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (227) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(1, 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U4(0)),F(0) -> U5^1(U4(0))) ---------------------------------------- (228) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U4(0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (229) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(U1(1), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(1, 0)),F(0) -> U5^1(U3(1, 0))) ---------------------------------------- (230) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(U1(not(0)), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(1, 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (231) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(U1(not(0)), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(U1(1), 0)),F(0) -> U5^1(U3(U1(1), 0))) ---------------------------------------- (232) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (233) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(1, 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(1))),F(0) -> U5^1(U1(not(1)))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(1, 0)),F(0) -> U5^1(U3(1, 0))) ---------------------------------------- (234) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(U1(1), 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U2(0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (235) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (236) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(implies(U1(1), 0)) U5^1(1) -> F(0) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (237) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(U1(1), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(U1(1)))),F(0) -> U5^1(U1(not(U1(1))))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(U1(1), 0)),F(0) -> U5^1(U3(U1(1), 0))) (F(0) -> U5^1(implies(1, 0)),F(0) -> U5^1(implies(1, 0))) ---------------------------------------- (238) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(U2(0)) F(0) -> U5^1(implies(1, 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (239) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (240) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(1))) U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(implies(1, 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (241) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(1))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(0)),F(0) -> U5^1(U1(0))) ---------------------------------------- (242) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(U1(1)))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(U1(0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (243) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (244) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(U1(1)))) U5^1(1) -> F(0) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(implies(1, 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (245) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U1(not(U1(1)))) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(1))),F(0) -> U5^1(U1(not(1)))) ---------------------------------------- (246) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(U1(not(1))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (247) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U4(0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(0),F(0) -> U5^1(0)) ---------------------------------------- (248) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(0) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (249) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (250) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U3(1, 0)) U5^1(1) -> F(0) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(U1(not(1))) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (251) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(1, 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U4(0)),F(0) -> U5^1(U4(0))) ---------------------------------------- (252) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U3(U1(1), 0)) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U4(0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (253) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(U3(U1(1), 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U3(1, 0)),F(0) -> U5^1(U3(1, 0))) ---------------------------------------- (254) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(implies(1, 0)) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(1, 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (255) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(0) -> U5^1(implies(1, 0)) at position [0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(not(1))),F(0) -> U5^1(U1(not(1)))) (F(0) -> U5^1(U2(0)),F(0) -> U5^1(U2(0))) (F(0) -> U5^1(U3(1, 0)),F(0) -> U5^1(U3(1, 0))) ---------------------------------------- (256) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(1))) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(1, 0)) F(0) -> U5^1(U2(0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (257) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (258) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(1))) U5^1(1) -> F(0) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(1, 0)) The TRS R consists of the following rules: implies(x, y) -> U1(not(x)) implies(x, y) -> U2(y) implies(x, y) -> U3(x, y) U3(1, y) -> U4(y) U4(0) -> 0 U2(1) -> 1 not(1) -> 0 not(0) -> 1 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (259) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (260) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(1))) U5^1(1) -> F(0) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(1, 0)) The TRS R consists of the following rules: U3(1, y) -> U4(y) U4(0) -> 0 not(1) -> 0 U1(1) -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (261) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (262) Obligation: Q DP problem: The TRS P consists of the following rules: F(0) -> U5^1(U1(not(1))) U5^1(1) -> F(0) F(0) -> U5^1(U4(0)) F(0) -> U5^1(U3(1, 0)) The TRS R consists of the following rules: U3(1, y) -> U4(y) U4(0) -> 0 not(1) -> 0 U1(1) -> 1 The set Q consists of the following terms: U3(1, x0) U4(0) not(1) U1(1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (263) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (264) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(1))) The TRS R consists of the following rules: U3(1, y) -> U4(y) U4(0) -> 0 not(1) -> 0 U1(1) -> 1 The set Q consists of the following terms: U3(1, x0) U4(0) not(1) U1(1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (265) 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. ---------------------------------------- (266) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(1))) The TRS R consists of the following rules: not(1) -> 0 U1(1) -> 1 The set Q consists of the following terms: U3(1, x0) U4(0) not(1) U1(1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (267) 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]. U3(1, x0) U4(0) ---------------------------------------- (268) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(not(1))) The TRS R consists of the following rules: not(1) -> 0 U1(1) -> 1 The set Q consists of the following terms: not(1) U1(1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (269) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(0) -> U5^1(U1(not(1))) at position [0,0] we obtained the following new rules [LPAR04]: (F(0) -> U5^1(U1(0)),F(0) -> U5^1(U1(0))) ---------------------------------------- (270) Obligation: Q DP problem: The TRS P consists of the following rules: U5^1(1) -> F(0) F(0) -> U5^1(U1(0)) The TRS R consists of the following rules: not(1) -> 0 U1(1) -> 1 The set Q consists of the following terms: not(1) U1(1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (271) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes. ---------------------------------------- (272) TRUE