NO proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QReductionProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QReductionProof [EQUIVALENT, 0 ms] (16) QDP (17) TransformationProof [EQUIVALENT, 0 ms] (18) QDP (19) UsableRulesProof [EQUIVALENT, 0 ms] (20) QDP (21) QReductionProof [EQUIVALENT, 0 ms] (22) QDP (23) TransformationProof [EQUIVALENT, 0 ms] (24) QDP (25) TransformationProof [EQUIVALENT, 0 ms] (26) QDP (27) TransformationProof [EQUIVALENT, 0 ms] (28) QDP (29) TransformationProof [EQUIVALENT, 0 ms] (30) QDP (31) TransformationProof [EQUIVALENT, 0 ms] (32) QDP (33) TransformationProof [EQUIVALENT, 0 ms] (34) QDP (35) TransformationProof [EQUIVALENT, 0 ms] (36) QDP (37) TransformationProof [EQUIVALENT, 1 ms] (38) QDP (39) TransformationProof [EQUIVALENT, 0 ms] (40) QDP (41) TransformationProof [EQUIVALENT, 0 ms] (42) QDP (43) TransformationProof [EQUIVALENT, 0 ms] (44) QDP (45) TransformationProof [EQUIVALENT, 0 ms] (46) QDP (47) TransformationProof [EQUIVALENT, 0 ms] (48) QDP (49) TransformationProof [EQUIVALENT, 0 ms] (50) QDP (51) TransformationProof [EQUIVALENT, 0 ms] (52) QDP (53) TransformationProof [EQUIVALENT, 0 ms] (54) QDP (55) TransformationProof [EQUIVALENT, 0 ms] (56) QDP (57) TransformationProof [EQUIVALENT, 0 ms] (58) QDP (59) TransformationProof [EQUIVALENT, 0 ms] (60) QDP (61) TransformationProof [EQUIVALENT, 0 ms] (62) QDP (63) TransformationProof [EQUIVALENT, 0 ms] (64) QDP (65) TransformationProof [EQUIVALENT, 0 ms] (66) QDP (67) TransformationProof [EQUIVALENT, 0 ms] (68) QDP (69) TransformationProof [EQUIVALENT, 0 ms] (70) QDP (71) TransformationProof [EQUIVALENT, 0 ms] (72) QDP (73) TransformationProof [EQUIVALENT, 0 ms] (74) QDP (75) TransformationProof [EQUIVALENT, 0 ms] (76) QDP (77) TransformationProof [EQUIVALENT, 0 ms] (78) QDP (79) TransformationProof [EQUIVALENT, 0 ms] (80) QDP (81) TransformationProof [EQUIVALENT, 0 ms] (82) QDP (83) TransformationProof [EQUIVALENT, 0 ms] (84) QDP (85) TransformationProof [EQUIVALENT, 0 ms] (86) QDP (87) TransformationProof [EQUIVALENT, 0 ms] (88) QDP (89) TransformationProof [EQUIVALENT, 0 ms] (90) QDP (91) TransformationProof [EQUIVALENT, 0 ms] (92) QDP (93) TransformationProof [EQUIVALENT, 0 ms] (94) QDP (95) TransformationProof [EQUIVALENT, 0 ms] (96) QDP (97) TransformationProof [EQUIVALENT, 0 ms] (98) QDP (99) TransformationProof [EQUIVALENT, 0 ms] (100) QDP (101) TransformationProof [EQUIVALENT, 0 ms] (102) QDP (103) TransformationProof [EQUIVALENT, 0 ms] (104) QDP (105) TransformationProof [EQUIVALENT, 0 ms] (106) QDP (107) TransformationProof [EQUIVALENT, 0 ms] (108) QDP (109) TransformationProof [EQUIVALENT, 0 ms] (110) QDP (111) TransformationProof [EQUIVALENT, 0 ms] (112) QDP (113) TransformationProof [EQUIVALENT, 0 ms] (114) QDP (115) TransformationProof [EQUIVALENT, 0 ms] (116) QDP (117) TransformationProof [EQUIVALENT, 0 ms] (118) QDP (119) TransformationProof [EQUIVALENT, 0 ms] (120) QDP (121) TransformationProof [EQUIVALENT, 0 ms] (122) QDP (123) MNOCProof [EQUIVALENT, 0 ms] (124) QDP (125) NonTerminationLoopProof [COMPLETE, 0 ms] (126) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: ack(0, y) -> s(y) ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) init -> ack(s(s(s(s(s(s(s(0))))))), 0) run_again(x) -> run_again(init) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) init run_again(x0) ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: ACK(s(x), 0) -> ACK(x, s(0)) ACK(s(x), s(y)) -> ACK(x, ack(s(x), y)) ACK(s(x), s(y)) -> ACK(s(x), y) INIT -> ACK(s(s(s(s(s(s(s(0))))))), 0) RUN_AGAIN(x) -> RUN_AGAIN(init) RUN_AGAIN(x) -> INIT The TRS R consists of the following rules: ack(0, y) -> s(y) ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) init -> ack(s(s(s(s(s(s(s(0))))))), 0) run_again(x) -> run_again(init) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) init run_again(x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: ACK(s(x), s(y)) -> ACK(x, ack(s(x), y)) ACK(s(x), s(y)) -> ACK(s(x), y) ACK(s(x), 0) -> ACK(x, s(0)) The TRS R consists of the following rules: ack(0, y) -> s(y) ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) init -> ack(s(s(s(s(s(s(s(0))))))), 0) run_again(x) -> run_again(init) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) init run_again(x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) 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. ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: ACK(s(x), s(y)) -> ACK(x, ack(s(x), y)) ACK(s(x), s(y)) -> ACK(s(x), y) ACK(s(x), 0) -> ACK(x, s(0)) The TRS R consists of the following rules: ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(s(x), 0) -> ack(x, s(0)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) init run_again(x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) 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]. init run_again(x0) ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: ACK(s(x), s(y)) -> ACK(x, ack(s(x), y)) ACK(s(x), s(y)) -> ACK(s(x), y) ACK(s(x), 0) -> ACK(x, s(0)) The TRS R consists of the following rules: ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(s(x), 0) -> ack(x, s(0)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *ACK(s(x), 0) -> ACK(x, s(0)) The graph contains the following edges 1 > 1 *ACK(s(x), s(y)) -> ACK(s(x), y) The graph contains the following edges 1 >= 1, 2 > 2 *ACK(s(x), s(y)) -> ACK(x, ack(s(x), y)) The graph contains the following edges 1 > 1 ---------------------------------------- (11) YES ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: RUN_AGAIN(x) -> RUN_AGAIN(init) The TRS R consists of the following rules: ack(0, y) -> s(y) ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) init -> ack(s(s(s(s(s(s(s(0))))))), 0) run_again(x) -> run_again(init) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) init run_again(x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) 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. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: RUN_AGAIN(x) -> RUN_AGAIN(init) The TRS R consists of the following rules: init -> ack(s(s(s(s(s(s(s(0))))))), 0) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(s(x), 0) -> ack(x, s(0)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) init run_again(x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) 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]. run_again(x0) ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: RUN_AGAIN(x) -> RUN_AGAIN(init) The TRS R consists of the following rules: init -> ack(s(s(s(s(s(s(s(0))))))), 0) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(s(x), 0) -> ack(x, s(0)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) init We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule RUN_AGAIN(x) -> RUN_AGAIN(init) at position [0] we obtained the following new rules [LPAR04]: (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))) ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(s(s(0))))))), 0)) The TRS R consists of the following rules: init -> ack(s(s(s(s(s(s(s(0))))))), 0) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(s(x), 0) -> ack(x, s(0)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) init We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) 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. ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(s(s(0))))))), 0)) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) init We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) 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]. init ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(s(s(0))))))), 0)) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) TransformationProof (EQUIVALENT) 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]: (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)))) ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(s(0)))))), s(0))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) TransformationProof (EQUIVALENT) 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]: (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)))) ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(s(s(0)))))), 0))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) TransformationProof (EQUIVALENT) 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]: (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))))) ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: RUN_AGAIN(x) -> RUN_AGAIN(ack(s(s(s(s(s(0))))), ack(s(s(s(s(s(0))))), s(0)))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) TransformationProof (EQUIVALENT) 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]: (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))))) ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) TransformationProof (EQUIVALENT) 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]: (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)))))) ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) TransformationProof (EQUIVALENT) 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]: (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)))))) ---------------------------------------- (34) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (35) TransformationProof (EQUIVALENT) 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]: (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))))))) ---------------------------------------- (36) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (37) TransformationProof (EQUIVALENT) 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]: (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))))))) ---------------------------------------- (38) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (39) TransformationProof (EQUIVALENT) 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]: (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)))))))) ---------------------------------------- (40) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (41) TransformationProof (EQUIVALENT) 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]: (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)))))))) ---------------------------------------- (42) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (43) TransformationProof (EQUIVALENT) 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]: (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))))))))) ---------------------------------------- (44) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (45) TransformationProof (EQUIVALENT) 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]: (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))))))))) ---------------------------------------- (46) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (47) TransformationProof (EQUIVALENT) 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]: (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))))))))) ---------------------------------------- (48) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (49) TransformationProof (EQUIVALENT) 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]: (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))))))))) ---------------------------------------- (50) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (51) TransformationProof (EQUIVALENT) 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]: (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))))))))) ---------------------------------------- (52) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (53) TransformationProof (EQUIVALENT) 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]: (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))))))))) ---------------------------------------- (54) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (55) TransformationProof (EQUIVALENT) 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]: (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)))))))))) ---------------------------------------- (56) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (57) TransformationProof (EQUIVALENT) 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]: (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)))))))))) ---------------------------------------- (58) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (59) TransformationProof (EQUIVALENT) 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]: (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)))))))))) ---------------------------------------- (60) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (61) TransformationProof (EQUIVALENT) 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]: (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)))))))))) ---------------------------------------- (62) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (63) TransformationProof (EQUIVALENT) 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]: (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)))))))))) ---------------------------------------- (64) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (65) TransformationProof (EQUIVALENT) 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]: (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)))))))))) ---------------------------------------- (66) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (67) TransformationProof (EQUIVALENT) 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]: (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)))))))))) ---------------------------------------- (68) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (69) TransformationProof (EQUIVALENT) 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]: (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)))))))))) ---------------------------------------- (70) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (71) TransformationProof (EQUIVALENT) 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]: (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))))))))))) ---------------------------------------- (72) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (73) TransformationProof (EQUIVALENT) 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]: (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))))))))))) ---------------------------------------- (74) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (75) TransformationProof (EQUIVALENT) 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]: (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))))))))))) ---------------------------------------- (76) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (77) TransformationProof (EQUIVALENT) 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]: (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))))))))))) ---------------------------------------- (78) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (79) TransformationProof (EQUIVALENT) 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]: (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)))))))))))) ---------------------------------------- (80) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (81) TransformationProof (EQUIVALENT) 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]: (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)))))))))))) ---------------------------------------- (82) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (83) TransformationProof (EQUIVALENT) 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]: (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)))))))))))) ---------------------------------------- (84) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (85) TransformationProof (EQUIVALENT) 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]: (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)))))))))))) ---------------------------------------- (86) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (87) TransformationProof (EQUIVALENT) 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]: (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)))))))))))) ---------------------------------------- (88) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (89) TransformationProof (EQUIVALENT) 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]: (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)))))))))))) ---------------------------------------- (90) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (91) TransformationProof (EQUIVALENT) 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]: (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)))))))))))) ---------------------------------------- (92) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (93) TransformationProof (EQUIVALENT) 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]: (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)))))))))))) ---------------------------------------- (94) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (95) TransformationProof (EQUIVALENT) 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]: (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)))))))))))) ---------------------------------------- (96) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (97) TransformationProof (EQUIVALENT) 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]: (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)))))))))))) ---------------------------------------- (98) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (99) TransformationProof (EQUIVALENT) 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]: (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)))))))))))) ---------------------------------------- (100) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (101) TransformationProof (EQUIVALENT) 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]: (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)))))))))))) ---------------------------------------- (102) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (103) TransformationProof (EQUIVALENT) 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]: (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)))))))))))) ---------------------------------------- (104) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (105) TransformationProof (EQUIVALENT) 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]: (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)))))))))))) ---------------------------------------- (106) Obligation: Q DP problem: The TRS P consists of the following rules: 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))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (107) TransformationProof (EQUIVALENT) 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]: (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))))))))))))) ---------------------------------------- (108) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (109) TransformationProof (EQUIVALENT) 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]: (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))))))))))))) ---------------------------------------- (110) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (111) TransformationProof (EQUIVALENT) 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]: (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))))))))))))) ---------------------------------------- (112) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (113) TransformationProof (EQUIVALENT) 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]: (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))))))))))))) ---------------------------------------- (114) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (115) TransformationProof (EQUIVALENT) 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]: (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))))))))))))) ---------------------------------------- (116) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (117) TransformationProof (EQUIVALENT) 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]: (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))))))))))))) ---------------------------------------- (118) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (119) TransformationProof (EQUIVALENT) 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]: (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))))))))))))) ---------------------------------------- (120) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (121) TransformationProof (EQUIVALENT) 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]: (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))))))))))))) ---------------------------------------- (122) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) The set Q consists of the following terms: ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (123) MNOCProof (EQUIVALENT) We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set. ---------------------------------------- (124) Obligation: Q DP problem: The TRS P consists of the following rules: 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)))))))))))) The TRS R consists of the following rules: ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(0, y) -> s(y) Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (125) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by semiunifying a rule from P directly. 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)))))))))))) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * 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)))))))))))] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence 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)))))))))))). ---------------------------------------- (126) NO