7.90/5.59 NO 7.90/5.60 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 7.90/5.60 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 7.90/5.60 7.90/5.60 7.90/5.60 Termination w.r.t. Q of the given QTRS could be disproven: 7.90/5.60 7.90/5.60 (0) QTRS 7.90/5.60 (1) DependencyPairsProof [EQUIVALENT, 0 ms] 7.90/5.60 (2) QDP 7.90/5.60 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 7.90/5.60 (4) AND 7.90/5.60 (5) QDP 7.90/5.60 (6) UsableRulesProof [EQUIVALENT, 0 ms] 7.90/5.60 (7) QDP 7.90/5.60 (8) QReductionProof [EQUIVALENT, 0 ms] 7.90/5.60 (9) QDP 7.90/5.60 (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] 7.90/5.60 (11) YES 7.90/5.60 (12) QDP 7.90/5.60 (13) UsableRulesProof [EQUIVALENT, 0 ms] 7.90/5.60 (14) QDP 7.90/5.60 (15) QReductionProof [EQUIVALENT, 0 ms] 7.90/5.60 (16) QDP 7.90/5.60 (17) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (18) QDP 7.90/5.60 (19) UsableRulesProof [EQUIVALENT, 0 ms] 7.90/5.60 (20) QDP 7.90/5.60 (21) QReductionProof [EQUIVALENT, 0 ms] 7.90/5.60 (22) QDP 7.90/5.60 (23) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (24) QDP 7.90/5.60 (25) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (26) QDP 7.90/5.60 (27) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (28) QDP 7.90/5.60 (29) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (30) QDP 7.90/5.60 (31) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (32) QDP 7.90/5.60 (33) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (34) QDP 7.90/5.60 (35) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (36) QDP 7.90/5.60 (37) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (38) QDP 7.90/5.60 (39) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (40) QDP 7.90/5.60 (41) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (42) QDP 7.90/5.60 (43) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (44) QDP 7.90/5.60 (45) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (46) QDP 7.90/5.60 (47) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (48) QDP 7.90/5.60 (49) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (50) QDP 7.90/5.60 (51) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (52) QDP 7.90/5.60 (53) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (54) QDP 7.90/5.60 (55) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (56) QDP 7.90/5.60 (57) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (58) QDP 7.90/5.60 (59) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (60) QDP 7.90/5.60 (61) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (62) QDP 7.90/5.60 (63) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (64) QDP 7.90/5.60 (65) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (66) QDP 7.90/5.60 (67) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (68) QDP 7.90/5.60 (69) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (70) QDP 7.90/5.60 (71) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (72) QDP 7.90/5.60 (73) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (74) QDP 7.90/5.60 (75) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (76) QDP 7.90/5.60 (77) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (78) QDP 7.90/5.60 (79) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (80) QDP 7.90/5.60 (81) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (82) QDP 7.90/5.60 (83) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.60 (84) QDP 7.90/5.60 (85) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (86) QDP 7.90/5.61 (87) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (88) QDP 7.90/5.61 (89) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (90) QDP 7.90/5.61 (91) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (92) QDP 7.90/5.61 (93) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (94) QDP 7.90/5.61 (95) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (96) QDP 7.90/5.61 (97) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (98) QDP 7.90/5.61 (99) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (100) QDP 7.90/5.61 (101) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (102) QDP 7.90/5.61 (103) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (104) QDP 7.90/5.61 (105) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (106) QDP 7.90/5.61 (107) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (108) QDP 7.90/5.61 (109) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (110) QDP 7.90/5.61 (111) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (112) QDP 7.90/5.61 (113) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (114) QDP 7.90/5.61 (115) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (116) QDP 7.90/5.61 (117) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (118) QDP 7.90/5.61 (119) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (120) QDP 7.90/5.61 (121) TransformationProof [EQUIVALENT, 0 ms] 7.90/5.61 (122) QDP 7.90/5.61 (123) MNOCProof [EQUIVALENT, 0 ms] 7.90/5.61 (124) QDP 7.90/5.61 (125) NonTerminationLoopProof [COMPLETE, 0 ms] 7.90/5.61 (126) NO 7.90/5.61 7.90/5.61 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (0) 7.90/5.61 Obligation: 7.90/5.61 Q restricted rewrite system: 7.90/5.61 The TRS R consists of the following rules: 7.90/5.61 7.90/5.61 ack(0, y) -> s(y) 7.90/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.90/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.90/5.61 init -> ack(s(s(s(s(s(s(s(0))))))), 0) 7.90/5.61 run_again(x) -> run_again(init) 7.90/5.61 7.90/5.61 The set Q consists of the following terms: 7.90/5.61 7.90/5.61 ack(0, x0) 7.90/5.61 ack(s(x0), 0) 7.90/5.61 ack(s(x0), s(x1)) 7.90/5.61 init 7.90/5.61 run_again(x0) 7.90/5.61 7.90/5.61 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (1) DependencyPairsProof (EQUIVALENT) 7.90/5.61 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (2) 7.90/5.61 Obligation: 7.90/5.61 Q DP problem: 7.90/5.61 The TRS P consists of the following rules: 7.90/5.61 7.90/5.61 ACK(s(x), 0) -> ACK(x, s(0)) 7.90/5.61 ACK(s(x), s(y)) -> ACK(x, ack(s(x), y)) 7.90/5.61 ACK(s(x), s(y)) -> ACK(s(x), y) 7.90/5.61 INIT -> ACK(s(s(s(s(s(s(s(0))))))), 0) 7.90/5.61 RUN_AGAIN(x) -> RUN_AGAIN(init) 7.90/5.61 RUN_AGAIN(x) -> INIT 7.90/5.61 7.90/5.61 The TRS R consists of the following rules: 7.90/5.61 7.90/5.61 ack(0, y) -> s(y) 7.90/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.90/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.90/5.61 init -> ack(s(s(s(s(s(s(s(0))))))), 0) 7.90/5.61 run_again(x) -> run_again(init) 7.90/5.61 7.90/5.61 The set Q consists of the following terms: 7.90/5.61 7.90/5.61 ack(0, x0) 7.90/5.61 ack(s(x0), 0) 7.90/5.61 ack(s(x0), s(x1)) 7.90/5.61 init 7.90/5.61 run_again(x0) 7.90/5.61 7.90/5.61 We have to consider all minimal (P,Q,R)-chains. 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (3) DependencyGraphProof (EQUIVALENT) 7.90/5.61 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes. 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (4) 7.90/5.61 Complex Obligation (AND) 7.90/5.61 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (5) 7.90/5.61 Obligation: 7.90/5.61 Q DP problem: 7.90/5.61 The TRS P consists of the following rules: 7.90/5.61 7.90/5.61 ACK(s(x), s(y)) -> ACK(x, ack(s(x), y)) 7.90/5.61 ACK(s(x), s(y)) -> ACK(s(x), y) 7.90/5.61 ACK(s(x), 0) -> ACK(x, s(0)) 7.90/5.61 7.90/5.61 The TRS R consists of the following rules: 7.90/5.61 7.90/5.61 ack(0, y) -> s(y) 7.90/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.90/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.90/5.61 init -> ack(s(s(s(s(s(s(s(0))))))), 0) 7.90/5.61 run_again(x) -> run_again(init) 7.90/5.61 7.90/5.61 The set Q consists of the following terms: 7.90/5.61 7.90/5.61 ack(0, x0) 7.90/5.61 ack(s(x0), 0) 7.90/5.61 ack(s(x0), s(x1)) 7.90/5.61 init 7.90/5.61 run_again(x0) 7.90/5.61 7.90/5.61 We have to consider all minimal (P,Q,R)-chains. 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (6) UsableRulesProof (EQUIVALENT) 7.90/5.61 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. 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (7) 7.90/5.61 Obligation: 7.90/5.61 Q DP problem: 7.90/5.61 The TRS P consists of the following rules: 7.90/5.61 7.90/5.61 ACK(s(x), s(y)) -> ACK(x, ack(s(x), y)) 7.90/5.61 ACK(s(x), s(y)) -> ACK(s(x), y) 7.90/5.61 ACK(s(x), 0) -> ACK(x, s(0)) 7.90/5.61 7.90/5.61 The TRS R consists of the following rules: 7.90/5.61 7.90/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.90/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.90/5.61 ack(0, y) -> s(y) 7.90/5.61 7.90/5.61 The set Q consists of the following terms: 7.90/5.61 7.90/5.61 ack(0, x0) 7.90/5.61 ack(s(x0), 0) 7.90/5.61 ack(s(x0), s(x1)) 7.90/5.61 init 7.90/5.61 run_again(x0) 7.90/5.61 7.90/5.61 We have to consider all minimal (P,Q,R)-chains. 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (8) QReductionProof (EQUIVALENT) 7.90/5.61 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 7.90/5.61 7.90/5.61 init 7.90/5.61 run_again(x0) 7.90/5.61 7.90/5.61 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (9) 7.90/5.61 Obligation: 7.90/5.61 Q DP problem: 7.90/5.61 The TRS P consists of the following rules: 7.90/5.61 7.90/5.61 ACK(s(x), s(y)) -> ACK(x, ack(s(x), y)) 7.90/5.61 ACK(s(x), s(y)) -> ACK(s(x), y) 7.90/5.61 ACK(s(x), 0) -> ACK(x, s(0)) 7.90/5.61 7.90/5.61 The TRS R consists of the following rules: 7.90/5.61 7.90/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.90/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.90/5.61 ack(0, y) -> s(y) 7.90/5.61 7.90/5.61 The set Q consists of the following terms: 7.90/5.61 7.90/5.61 ack(0, x0) 7.90/5.61 ack(s(x0), 0) 7.90/5.61 ack(s(x0), s(x1)) 7.90/5.61 7.90/5.61 We have to consider all minimal (P,Q,R)-chains. 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (10) QDPSizeChangeProof (EQUIVALENT) 7.90/5.61 By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 7.90/5.61 7.90/5.61 From the DPs we obtained the following set of size-change graphs: 7.90/5.61 *ACK(s(x), 0) -> ACK(x, s(0)) 7.90/5.61 The graph contains the following edges 1 > 1 7.90/5.61 7.90/5.61 7.90/5.61 *ACK(s(x), s(y)) -> ACK(s(x), y) 7.90/5.61 The graph contains the following edges 1 >= 1, 2 > 2 7.90/5.61 7.90/5.61 7.90/5.61 *ACK(s(x), s(y)) -> ACK(x, ack(s(x), y)) 7.90/5.61 The graph contains the following edges 1 > 1 7.90/5.61 7.90/5.61 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (11) 7.90/5.61 YES 7.90/5.61 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (12) 7.90/5.61 Obligation: 7.90/5.61 Q DP problem: 7.90/5.61 The TRS P consists of the following rules: 7.90/5.61 7.90/5.61 RUN_AGAIN(x) -> RUN_AGAIN(init) 7.90/5.61 7.90/5.61 The TRS R consists of the following rules: 7.90/5.61 7.90/5.61 ack(0, y) -> s(y) 7.90/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.90/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.90/5.61 init -> ack(s(s(s(s(s(s(s(0))))))), 0) 7.90/5.61 run_again(x) -> run_again(init) 7.90/5.61 7.90/5.61 The set Q consists of the following terms: 7.90/5.61 7.90/5.61 ack(0, x0) 7.90/5.61 ack(s(x0), 0) 7.90/5.61 ack(s(x0), s(x1)) 7.90/5.61 init 7.90/5.61 run_again(x0) 7.90/5.61 7.90/5.61 We have to consider all minimal (P,Q,R)-chains. 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (13) UsableRulesProof (EQUIVALENT) 7.90/5.61 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. 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (14) 7.90/5.61 Obligation: 7.90/5.61 Q DP problem: 7.90/5.61 The TRS P consists of the following rules: 7.90/5.61 7.90/5.61 RUN_AGAIN(x) -> RUN_AGAIN(init) 7.90/5.61 7.90/5.61 The TRS R consists of the following rules: 7.90/5.61 7.90/5.61 init -> ack(s(s(s(s(s(s(s(0))))))), 0) 7.90/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.90/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.90/5.61 ack(0, y) -> s(y) 7.90/5.61 7.90/5.61 The set Q consists of the following terms: 7.90/5.61 7.90/5.61 ack(0, x0) 7.90/5.61 ack(s(x0), 0) 7.90/5.61 ack(s(x0), s(x1)) 7.90/5.61 init 7.90/5.61 run_again(x0) 7.90/5.61 7.90/5.61 We have to consider all minimal (P,Q,R)-chains. 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (15) QReductionProof (EQUIVALENT) 7.90/5.61 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 7.90/5.61 7.90/5.61 run_again(x0) 7.90/5.61 7.90/5.61 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (16) 7.90/5.61 Obligation: 7.90/5.61 Q DP problem: 7.90/5.61 The TRS P consists of the following rules: 7.90/5.61 7.90/5.61 RUN_AGAIN(x) -> RUN_AGAIN(init) 7.90/5.61 7.90/5.61 The TRS R consists of the following rules: 7.90/5.61 7.90/5.61 init -> ack(s(s(s(s(s(s(s(0))))))), 0) 7.90/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.90/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.90/5.61 ack(0, y) -> s(y) 7.90/5.61 7.90/5.61 The set Q consists of the following terms: 7.90/5.61 7.90/5.61 ack(0, x0) 7.90/5.61 ack(s(x0), 0) 7.90/5.61 ack(s(x0), s(x1)) 7.90/5.61 init 7.90/5.61 7.90/5.61 We have to consider all minimal (P,Q,R)-chains. 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (17) TransformationProof (EQUIVALENT) 7.90/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(init) at position [0] we obtained the following new rules [LPAR04]: 7.90/5.61 7.90/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(s(s(0))))))), 0)),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(s(s(0))))))), 0))) 7.90/5.61 7.90/5.61 7.90/5.61 ---------------------------------------- 7.90/5.61 7.90/5.61 (18) 7.90/5.61 Obligation: 7.90/5.61 Q DP problem: 7.90/5.61 The TRS P consists of the following rules: 7.90/5.61 7.90/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(s(s(0))))))), 0)) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 init -> ack(s(s(s(s(s(s(s(0))))))), 0) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 init 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (19) UsableRulesProof (EQUIVALENT) 7.97/5.61 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. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (20) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(s(s(0))))))), 0)) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 init 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (21) QReductionProof (EQUIVALENT) 7.97/5.61 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 7.97/5.61 7.97/5.61 init 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (22) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(s(s(0))))))), 0)) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (23) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(s(s(0))))))), 0)) at position [0] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(s(0)))))), s(0))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(s(0)))))), s(0)))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (24) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(s(0)))))), s(0))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (25) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(s(0)))))), s(0))) at position [0] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(s(s(0)))))), 0))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(s(s(0)))))), 0)))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (26) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(s(s(0)))))), 0))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (27) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(s(s(0)))))), 0))) at position [0,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(s(0))))), s(0)))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(s(0))))), s(0))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (28) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(s(0))))), s(0)))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (29) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(s(0))))), s(0)))) at position [0,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(s(s(0))))), 0)))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(s(s(0))))), 0))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (30) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(s(s(0))))), 0)))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (31) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(s(s(0))))), 0)))) at position [0,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(s(0)))), s(0))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(s(0)))), s(0)))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (32) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(s(0)))), s(0))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (33) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(s(0)))), s(0))))) at position [0,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(s(0)))), 0))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(s(0)))), 0)))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (34) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(s(0)))), 0))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (35) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(s(0)))), 0))))) at position [0,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(0))), s(0)))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(0))), s(0))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (36) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(0))), s(0)))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (37) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(0))), s(0)))))) at position [0,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(s(0))), 0)))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(s(0))), 0))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (38) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(s(0))), 0)))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (39) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(s(0))), 0)))))) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(0)), s(0))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(0)), s(0)))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (40) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(0)), s(0))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (41) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(0)), s(0))))))) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(s(0)), 0))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(s(0)), 0)))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (42) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(s(0)), 0))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (43) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(s(0)), 0))))))) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(0), s(0)))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(0), s(0))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (44) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(0), s(0)))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (45) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(0), s(0)))))))) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(0, ack(s(0), 0)))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(0, ack(s(0), 0))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (46) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(0, ack(s(0), 0)))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (47) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(0, ack(s(0), 0)))))))) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), s(ack(s(0), 0)))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), s(ack(s(0), 0))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (48) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), s(ack(s(0), 0)))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (49) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), s(ack(s(0), 0)))))))) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(0, ack(s(0), ack(s(0), 0)))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(0, ack(s(0), ack(s(0), 0))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (50) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(0, ack(s(0), ack(s(0), 0)))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (51) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(0, ack(s(0), ack(s(0), 0)))))))) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), s(ack(s(0), ack(s(0), 0)))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), s(ack(s(0), ack(s(0), 0))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (52) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), s(ack(s(0), ack(s(0), 0)))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (53) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), s(ack(s(0), ack(s(0), 0)))))))) at position [0,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), ack(s(0), 0)))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), ack(s(0), 0))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (54) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), ack(s(0), 0)))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (55) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), ack(s(0), 0)))))))) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), ack(0, s(0))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), ack(0, s(0)))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (56) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), ack(0, s(0))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (57) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), ack(0, s(0))))))))) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), s(s(0))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), s(s(0)))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (58) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), s(s(0))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (59) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), s(s(0))))))))) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(0, ack(s(0), s(0))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(0, ack(s(0), s(0)))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (60) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(0, ack(s(0), s(0))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (61) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(0, ack(s(0), s(0))))))))) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), s(ack(s(0), s(0))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), s(ack(s(0), s(0)))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (62) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), s(ack(s(0), s(0))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (63) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), s(ack(s(0), s(0))))))))) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), ack(s(0), s(0))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), ack(s(0), s(0)))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (64) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), ack(s(0), s(0))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (65) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), ack(s(0), s(0))))))))) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), ack(0, ack(s(0), 0))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), ack(0, ack(s(0), 0)))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (66) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), ack(0, ack(s(0), 0))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (67) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), ack(0, ack(s(0), 0))))))))) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), s(ack(s(0), 0))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), s(ack(s(0), 0)))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (68) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), s(ack(s(0), 0))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (69) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), s(ack(s(0), 0))))))))) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), ack(s(0), 0))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), ack(s(0), 0)))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (70) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), ack(s(0), 0))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (71) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), ack(s(0), 0))))))))) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), ack(0, s(0)))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), ack(0, s(0))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (72) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), ack(0, s(0)))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (73) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), ack(0, s(0)))))))))) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), s(s(0)))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), s(s(0))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (74) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), s(s(0)))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (75) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), s(s(0)))))))))) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), s(0)))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), s(0))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (76) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), s(0)))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (77) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), s(0)))))))))) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), 0)))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), 0))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (78) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), 0)))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (79) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), 0)))))))))) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(0))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(0)))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (80) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(0))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (81) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(0))))))))))) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), 0))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (82) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), 0))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (83) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), 0))))))))))) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), 0))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (84) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), 0))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (85) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), 0))))))))))) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), 0))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (86) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), 0))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (87) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), 0))))))))))) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), 0))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (88) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), 0))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (89) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), 0))))))))))) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), 0))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (90) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), 0))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (91) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), 0))))))))))) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), 0))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (92) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), 0))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (93) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), 0))))))))))) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (94) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (95) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (96) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (97) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (98) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (99) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (100) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (101) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))) at position [0,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (102) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (103) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))) at position [0,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (104) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (105) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(s(0))), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))) at position [0,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (106) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (107) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0))))))))))) at position [0,1,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, s(0)))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, s(0))))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (108) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, s(0)))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (109) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, s(0)))))))))))) at position [0,1,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(s(0)))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(s(0))))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (110) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(s(0)))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (111) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(s(0)))))))))))) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), s(0)))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), s(0))))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (112) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), s(0)))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (113) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), s(0)))))))))))) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), s(0)))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), s(0))))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (114) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), s(0)))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (115) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), s(0)))))))))))) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), s(0)))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), s(0))))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (116) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), s(0)))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (117) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), s(0)))))))))))) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), s(0)))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), s(0))))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (118) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), s(0)))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (119) TransformationProof (EQUIVALENT) 7.97/5.61 By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), s(0)))))))))))) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), s(0)))))))))))),RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), s(0))))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (120) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), s(0)))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (121) TransformationProof (EQUIVALENT) 7.97/5.61 By narrowing [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), s(0)))))))))))) at position [0,1,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: 7.97/5.61 7.97/5.61 (RUN_AGAIN(y0) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(0, ack(s(0), 0)))))))))))),RUN_AGAIN(y0) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(0, ack(s(0), 0))))))))))))) 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (122) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(y0) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(0, ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 The set Q consists of the following terms: 7.97/5.61 7.97/5.61 ack(0, x0) 7.97/5.61 ack(s(x0), 0) 7.97/5.61 ack(s(x0), s(x1)) 7.97/5.61 7.97/5.61 We have to consider all minimal (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (123) MNOCProof (EQUIVALENT) 7.97/5.61 We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (124) 7.97/5.61 Obligation: 7.97/5.61 Q DP problem: 7.97/5.61 The TRS P consists of the following rules: 7.97/5.61 7.97/5.61 RUN_AGAIN(y0) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(0, ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 The TRS R consists of the following rules: 7.97/5.61 7.97/5.61 ack(s(x), 0) -> ack(x, s(0)) 7.97/5.61 ack(s(x), s(y)) -> ack(x, ack(s(x), y)) 7.97/5.61 ack(0, y) -> s(y) 7.97/5.61 7.97/5.61 Q is empty. 7.97/5.61 We have to consider all (P,Q,R)-chains. 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (125) NonTerminationLoopProof (COMPLETE) 7.97/5.61 We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. 7.97/5.61 Found a loop by semiunifying a rule from P directly. 7.97/5.61 7.97/5.61 s = RUN_AGAIN(y0) evaluates to t =RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(0, ack(s(0), 0)))))))))))) 7.97/5.61 7.97/5.61 Thus s starts an infinite chain as s semiunifies with t with the following substitutions: 7.97/5.61 * Matcher: [y0 / ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(0, ack(s(0), 0)))))))))))] 7.97/5.61 * Semiunifier: [ ] 7.97/5.61 7.97/5.61 -------------------------------------------------------------------------------- 7.97/5.61 Rewriting sequence 7.97/5.61 7.97/5.61 The DP semiunifies directly so there is only one rewrite step from RUN_AGAIN(y0) to RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(0, ack(s(0), 0)))))))))))). 7.97/5.61 7.97/5.61 7.97/5.61 7.97/5.61 7.97/5.61 ---------------------------------------- 7.97/5.61 7.97/5.61 (126) 7.97/5.61 NO 7.99/6.43 EOF