YES 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 3 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES (21) QDP (22) UsableRulesProof [EQUIVALENT, 0 ms] (23) QDP (24) QReductionProof [EQUIVALENT, 0 ms] (25) QDP (26) TransformationProof [EQUIVALENT, 0 ms] (27) QDP (28) UsableRulesProof [EQUIVALENT, 0 ms] (29) QDP (30) QReductionProof [EQUIVALENT, 0 ms] (31) QDP (32) TransformationProof [EQUIVALENT, 0 ms] (33) QDP (34) TransformationProof [EQUIVALENT, 0 ms] (35) QDP (36) TransformationProof [EQUIVALENT, 0 ms] (37) QDP (38) TransformationProof [EQUIVALENT, 0 ms] (39) QDP (40) TransformationProof [EQUIVALENT, 0 ms] (41) QDP (42) TransformationProof [EQUIVALENT, 0 ms] (43) QDP (44) TransformationProof [EQUIVALENT, 0 ms] (45) QDP (46) TransformationProof [EQUIVALENT, 0 ms] (47) QDP (48) TransformationProof [EQUIVALENT, 0 ms] (49) QDP (50) TransformationProof [EQUIVALENT, 0 ms] (51) QDP (52) TransformationProof [EQUIVALENT, 0 ms] (53) QDP (54) TransformationProof [EQUIVALENT, 0 ms] (55) QDP (56) TransformationProof [EQUIVALENT, 0 ms] (57) QDP (58) TransformationProof [EQUIVALENT, 0 ms] (59) QDP (60) TransformationProof [EQUIVALENT, 0 ms] (61) QDP (62) TransformationProof [EQUIVALENT, 0 ms] (63) QDP (64) TransformationProof [EQUIVALENT, 0 ms] (65) QDP (66) TransformationProof [EQUIVALENT, 0 ms] (67) QDP (68) TransformationProof [EQUIVALENT, 0 ms] (69) QDP (70) TransformationProof [EQUIVALENT, 0 ms] (71) QDP (72) TransformationProof [EQUIVALENT, 0 ms] (73) QDP (74) TransformationProof [EQUIVALENT, 0 ms] (75) QDP (76) TransformationProof [EQUIVALENT, 1 ms] (77) QDP (78) TransformationProof [EQUIVALENT, 0 ms] (79) QDP (80) TransformationProof [EQUIVALENT, 0 ms] (81) QDP (82) TransformationProof [EQUIVALENT, 0 ms] (83) QDP (84) TransformationProof [EQUIVALENT, 0 ms] (85) QDP (86) TransformationProof [EQUIVALENT, 0 ms] (87) QDP (88) TransformationProof [EQUIVALENT, 0 ms] (89) QDP (90) TransformationProof [EQUIVALENT, 0 ms] (91) QDP (92) TransformationProof [EQUIVALENT, 0 ms] (93) QDP (94) TransformationProof [EQUIVALENT, 0 ms] (95) QDP (96) TransformationProof [EQUIVALENT, 0 ms] (97) QDP (98) TransformationProof [EQUIVALENT, 0 ms] (99) QDP (100) TransformationProof [EQUIVALENT, 0 ms] (101) QDP (102) TransformationProof [EQUIVALENT, 0 ms] (103) QDP (104) TransformationProof [EQUIVALENT, 0 ms] (105) QDP (106) TransformationProof [EQUIVALENT, 0 ms] (107) QDP (108) TransformationProof [EQUIVALENT, 0 ms] (109) QDP (110) TransformationProof [EQUIVALENT, 0 ms] (111) QDP (112) TransformationProof [EQUIVALENT, 0 ms] (113) QDP (114) TransformationProof [EQUIVALENT, 0 ms] (115) QDP (116) TransformationProof [EQUIVALENT, 0 ms] (117) QDP (118) TransformationProof [EQUIVALENT, 0 ms] (119) QDP (120) TransformationProof [EQUIVALENT, 0 ms] (121) QDP (122) TransformationProof [EQUIVALENT, 0 ms] (123) QDP (124) TransformationProof [EQUIVALENT, 0 ms] (125) QDP (126) TransformationProof [EQUIVALENT, 0 ms] (127) QDP (128) TransformationProof [EQUIVALENT, 0 ms] (129) QDP (130) RemovalProof [SOUND, 0 ms] (131) QDP (132) NonInfProof [EQUIVALENT, 40 ms] (133) QDP (134) DependencyGraphProof [EQUIVALENT, 0 ms] (135) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: numbers -> d(0) d(x) -> if(le(x, nr), x) if(true, x) -> cons(x, d(s(x))) if(false, x) -> nil le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) nr -> ack(s(s(s(s(s(s(0)))))), 0) ack(0, x) -> s(x) ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: numbers -> d(0) d(x) -> if(le(x, nr), x) if(true, x) -> cons(x, d(s(x))) if(false, x) -> nil le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) nr -> ack(s(s(s(s(s(s(0)))))), 0) ack(0, x) -> s(x) ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) The set Q consists of the following terms: numbers d(x0) if(true, x0) if(false, x0) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) nr ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: NUMBERS -> D(0) D(x) -> IF(le(x, nr), x) D(x) -> LE(x, nr) D(x) -> NR IF(true, x) -> D(s(x)) LE(s(x), s(y)) -> LE(x, y) NR -> ACK(s(s(s(s(s(s(0)))))), 0) 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) The TRS R consists of the following rules: numbers -> d(0) d(x) -> if(le(x, nr), x) if(true, x) -> cons(x, d(s(x))) if(false, x) -> nil le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) nr -> ack(s(s(s(s(s(s(0)))))), 0) ack(0, x) -> s(x) ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) The set Q consists of the following terms: numbers d(x0) if(true, x0) if(false, x0) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) nr ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 4 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (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: numbers -> d(0) d(x) -> if(le(x, nr), x) if(true, x) -> cons(x, d(s(x))) if(false, x) -> nil le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) nr -> ack(s(s(s(s(s(s(0)))))), 0) ack(0, x) -> s(x) ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) The set Q consists of the following terms: numbers d(x0) if(true, x0) if(false, x0) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) nr ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: 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, x) -> s(x) The set Q consists of the following terms: numbers d(x0) if(true, x0) if(false, x0) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) nr ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. numbers d(x0) if(true, x0) if(false, x0) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) nr ---------------------------------------- (11) 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, x) -> s(x) 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. ---------------------------------------- (12) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *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 ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: LE(s(x), s(y)) -> LE(x, y) The TRS R consists of the following rules: numbers -> d(0) d(x) -> if(le(x, nr), x) if(true, x) -> cons(x, d(s(x))) if(false, x) -> nil le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) nr -> ack(s(s(s(s(s(s(0)))))), 0) ack(0, x) -> s(x) ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) The set Q consists of the following terms: numbers d(x0) if(true, x0) if(false, x0) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) nr ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: LE(s(x), s(y)) -> LE(x, y) R is empty. The set Q consists of the following terms: numbers d(x0) if(true, x0) if(false, x0) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) nr ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. numbers d(x0) if(true, x0) if(false, x0) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) nr ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: LE(s(x), s(y)) -> LE(x, y) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *LE(s(x), s(y)) -> LE(x, y) The graph contains the following edges 1 > 1, 2 > 2 ---------------------------------------- (20) YES ---------------------------------------- (21) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, nr), x) The TRS R consists of the following rules: numbers -> d(0) d(x) -> if(le(x, nr), x) if(true, x) -> cons(x, d(s(x))) if(false, x) -> nil le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) nr -> ack(s(s(s(s(s(s(0)))))), 0) ack(0, x) -> s(x) ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) The set Q consists of the following terms: numbers d(x0) if(true, x0) if(false, x0) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) nr ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (22) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (23) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, nr), x) The TRS R consists of the following rules: nr -> ack(s(s(s(s(s(s(0)))))), 0) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(s(x), 0) -> ack(x, s(0)) ack(0, x) -> s(x) The set Q consists of the following terms: numbers d(x0) if(true, x0) if(false, x0) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) nr ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (24) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. numbers d(x0) if(true, x0) if(false, x0) ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, nr), x) The TRS R consists of the following rules: nr -> ack(s(s(s(s(s(s(0)))))), 0) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(s(x), 0) -> ack(x, s(0)) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) nr ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (26) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, nr), x) at position [0,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(s(s(0)))))), 0)), x),D(x) -> IF(le(x, ack(s(s(s(s(s(s(0)))))), 0)), x)) ---------------------------------------- (27) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(s(s(0)))))), 0)), x) The TRS R consists of the following rules: nr -> ack(s(s(s(s(s(s(0)))))), 0) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(s(x), 0) -> ack(x, s(0)) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) nr ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (28) 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. ---------------------------------------- (29) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(s(s(0)))))), 0)), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) nr ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (30) 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]. nr ---------------------------------------- (31) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(s(s(0)))))), 0)), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (32) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(s(s(0)))))), 0)), x) at position [0,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(s(0))))), s(0))), x),D(x) -> IF(le(x, ack(s(s(s(s(s(0))))), s(0))), x)) ---------------------------------------- (33) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(s(0))))), s(0))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (34) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(s(0))))), s(0))), x) at position [0,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(s(s(0))))), 0))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(s(s(0))))), 0))), x)) ---------------------------------------- (35) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(s(s(0))))), 0))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (36) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(s(s(0))))), 0))), x) at position [0,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(s(0)))), s(0)))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(s(0)))), s(0)))), x)) ---------------------------------------- (37) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(s(0)))), s(0)))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (38) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(s(0)))), s(0)))), x) at position [0,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(s(0)))), 0)))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(s(0)))), 0)))), x)) ---------------------------------------- (39) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(s(0)))), 0)))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (40) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(s(0)))), 0)))), x) at position [0,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(0))), s(0))))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(0))), s(0))))), x)) ---------------------------------------- (41) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(0))), s(0))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (42) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(0))), s(0))))), x) at position [0,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(s(0))), 0))))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(s(0))), 0))))), x)) ---------------------------------------- (43) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(s(0))), 0))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (44) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(s(0))), 0))))), x) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(0)), s(0)))))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(0)), s(0)))))), x)) ---------------------------------------- (45) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(0)), s(0)))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (46) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(0)), s(0)))))), x) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(s(0)), 0)))))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(s(0)), 0)))))), x)) ---------------------------------------- (47) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(s(0)), 0)))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (48) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(s(0)), 0)))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(0), s(0))))))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(0), s(0))))))), x)) ---------------------------------------- (49) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(0), s(0))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (50) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(0), s(0))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(0, ack(s(0), 0))))))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(0, ack(s(0), 0))))))), x)) ---------------------------------------- (51) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(0, ack(s(0), 0))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (52) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(0, ack(s(0), 0))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), s(ack(s(0), 0))))))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), s(ack(s(0), 0))))))), x)) ---------------------------------------- (53) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), s(ack(s(0), 0))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (54) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), s(ack(s(0), 0))))))), x) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(0, ack(s(0), ack(s(0), 0))))))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(0, ack(s(0), ack(s(0), 0))))))), x)) ---------------------------------------- (55) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(0, ack(s(0), ack(s(0), 0))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (56) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(0, ack(s(0), ack(s(0), 0))))))), x) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), s(ack(s(0), ack(s(0), 0))))))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), s(ack(s(0), ack(s(0), 0))))))), x)) ---------------------------------------- (57) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), s(ack(s(0), ack(s(0), 0))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (58) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), s(ack(s(0), ack(s(0), 0))))))), x) at position [0,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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))))))), x),D(x) -> IF(le(x, 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))))))), x)) ---------------------------------------- (59) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (60) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))), x),D(x) -> IF(le(x, 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)))))))), x)) ---------------------------------------- (61) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (62) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), s(s(0)))))))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), s(s(0)))))))), x)) ---------------------------------------- (63) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), s(s(0)))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (64) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), s(s(0)))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))), x),D(x) -> IF(le(x, 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)))))))), x)) ---------------------------------------- (65) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (66) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), s(ack(s(0), s(0)))))))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), s(ack(s(0), s(0)))))))), x)) ---------------------------------------- (67) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), s(ack(s(0), s(0)))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (68) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), s(ack(s(0), s(0)))))))), x) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))), x),D(x) -> IF(le(x, 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)))))))), x)) ---------------------------------------- (69) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (70) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))), x),D(x) -> IF(le(x, 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)))))))), x)) ---------------------------------------- (71) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (72) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))), x),D(x) -> IF(le(x, 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)))))))), x)) ---------------------------------------- (73) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (74) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))), x),D(x) -> IF(le(x, 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)))))))), x)) ---------------------------------------- (75) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (76) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))), x) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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))))))))), x),D(x) -> IF(le(x, 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))))))))), x)) ---------------------------------------- (77) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (78) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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))))))))), x) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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))))))))), x),D(x) -> IF(le(x, 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))))))))), x)) ---------------------------------------- (79) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (80) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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))))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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))))))))), x),D(x) -> IF(le(x, 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))))))))), x)) ---------------------------------------- (81) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (82) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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))))))))), x) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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))))))))), x),D(x) -> IF(le(x, 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))))))))), x)) ---------------------------------------- (83) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (84) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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))))))))), x) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))))), x),D(x) -> IF(le(x, 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)))))))))), x)) ---------------------------------------- (85) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (86) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))))), x) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))))), x),D(x) -> IF(le(x, 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)))))))))), x)) ---------------------------------------- (87) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (88) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))))), x) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))))), x),D(x) -> IF(le(x, 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)))))))))), x)) ---------------------------------------- (89) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (90) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))))), x) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))))), x),D(x) -> IF(le(x, 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)))))))))), x)) ---------------------------------------- (91) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (92) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))))), x) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))))), x),D(x) -> IF(le(x, 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)))))))))), x)) ---------------------------------------- (93) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (94) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))))), x),D(x) -> IF(le(x, 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)))))))))), x)) ---------------------------------------- (95) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (96) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))))), x),D(x) -> IF(le(x, 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)))))))))), x)) ---------------------------------------- (97) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (98) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))))), x),D(x) -> IF(le(x, 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)))))))))), x)) ---------------------------------------- (99) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (100) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))))), x),D(x) -> IF(le(x, 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)))))))))), x)) ---------------------------------------- (101) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (102) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))))), x) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))))), x),D(x) -> IF(le(x, 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)))))))))), x)) ---------------------------------------- (103) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (104) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))))), x) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))))), x),D(x) -> IF(le(x, 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)))))))))), x)) ---------------------------------------- (105) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (106) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))))), x) at position [0,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))))), x),D(x) -> IF(le(x, 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)))))))))), x)) ---------------------------------------- (107) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (108) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))))), x) at position [0,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))))), x),D(x) -> IF(le(x, 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)))))))))), x)) ---------------------------------------- (109) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (110) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))))), x) at position [0,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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)))))))))), x),D(x) -> IF(le(x, 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)))))))))), x)) ---------------------------------------- (111) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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)))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (112) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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)))))))))), x) at position [0,1,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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))))))))))), x),D(x) -> IF(le(x, 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))))))))))), x)) ---------------------------------------- (113) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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))))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (114) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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))))))))))), x) at position [0,1,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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))))))))))), x),D(x) -> IF(le(x, 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))))))))))), x)) ---------------------------------------- (115) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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))))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (116) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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))))))))))), x) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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))))))))))), x),D(x) -> IF(le(x, 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))))))))))), x)) ---------------------------------------- (117) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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))))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (118) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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))))))))))), x) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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))))))))))), x),D(x) -> IF(le(x, 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))))))))))), x)) ---------------------------------------- (119) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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))))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (120) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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))))))))))), x) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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))))))))))), x),D(x) -> IF(le(x, 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))))))))))), x)) ---------------------------------------- (121) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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))))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (122) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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))))))))))), x) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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))))))))))), x),D(x) -> IF(le(x, 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))))))))))), x)) ---------------------------------------- (123) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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))))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (124) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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))))))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, 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))))))))))), x),D(x) -> IF(le(x, 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))))))))))), x)) ---------------------------------------- (125) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, 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))))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (126) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, 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))))))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), x)) ---------------------------------------- (127) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (128) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]: (D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(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), s(0))))))))))), x),D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(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), s(0))))))))))), x)) ---------------------------------------- (129) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> D(s(x)) D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(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), s(0))))))))))), x) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (130) RemovalProof (SOUND) In the following pairs the term without variables ack(s(s(s(s(0)))), ack(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), s(0)))))))))) is replaced by the fresh variable x_removed. Pair: D(x) -> IF(le(x, ack(s(s(s(s(0)))), ack(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), s(0))))))))))), x) Positions in right side of the pair: *[0,1]The following rules were checked for non-overlappingness: ack(s(x), s(y)) -> ack(x, ack(s(x), y)) ack(s(x), 0) -> ack(x, s(0)) ack(0, x) -> s(x) The new variable was added to all pairs as a new argument[CONREM]. ---------------------------------------- (131) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x, x_removed) -> D(s(x), x_removed) D(x, x_removed) -> IF(le(x, x_removed), x, x_removed) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (132) NonInfProof (EQUIVALENT) The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: Note that final constraints are written in bold face. For Pair IF(true, x, x_removed) -> D(s(x), x_removed) the following chains were created: *We consider the chain D(x2, x3) -> IF(le(x2, x3), x2, x3), IF(true, x4, x5) -> D(s(x4), x5) which results in the following constraint: (1) (IF(le(x2, x3), x2, x3)=IF(true, x4, x5) ==> IF(true, x4, x5)_>=_D(s(x4), x5)) We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint: (2) (le(x2, x3)=true ==> IF(true, x2, x3)_>=_D(s(x2), x3)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on le(x2, x3)=true which results in the following new constraints: (3) (true=true ==> IF(true, 0, x12)_>=_D(s(0), x12)) (4) (le(x15, x14)=true & (le(x15, x14)=true ==> IF(true, x15, x14)_>=_D(s(x15), x14)) ==> IF(true, s(x15), s(x14))_>=_D(s(s(x15)), s(x14))) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (5) (IF(true, 0, x12)_>=_D(s(0), x12)) We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (le(x15, x14)=true ==> IF(true, x15, x14)_>=_D(s(x15), x14)) with sigma = [ ] which results in the following new constraint: (6) (IF(true, x15, x14)_>=_D(s(x15), x14) ==> IF(true, s(x15), s(x14))_>=_D(s(s(x15)), s(x14))) For Pair D(x, x_removed) -> IF(le(x, x_removed), x, x_removed) the following chains were created: *We consider the chain IF(true, x6, x7) -> D(s(x6), x7), D(x8, x9) -> IF(le(x8, x9), x8, x9) which results in the following constraint: (1) (D(s(x6), x7)=D(x8, x9) ==> D(x8, x9)_>=_IF(le(x8, x9), x8, x9)) We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint: (2) (D(s(x6), x7)_>=_IF(le(s(x6), x7), s(x6), x7)) To summarize, we get the following constraints P__>=_ for the following pairs. *IF(true, x, x_removed) -> D(s(x), x_removed) *(IF(true, 0, x12)_>=_D(s(0), x12)) *(IF(true, x15, x14)_>=_D(s(x15), x14) ==> IF(true, s(x15), s(x14))_>=_D(s(s(x15)), s(x14))) *D(x, x_removed) -> IF(le(x, x_removed), x, x_removed) *(D(s(x6), x7)_>=_IF(le(s(x6), x7), s(x6), x7)) The constraints for P_> respective P_bound are constructed from P__>=_ where we just replace every occurence of "t _>=_ s" in P__>=_ by "t > s" respective "t _>=_ c". Here c stands for the fresh constant used for P_bound. Using the following integer polynomial ordering the resulting constraints can be solved Polynomial interpretation [NONINF]: POL(0) = 0 POL(D(x_1, x_2)) = -1 - x_1 + x_2 POL(IF(x_1, x_2, x_3)) = -x_1 - x_2 + x_3 POL(c) = -2 POL(false) = 1 POL(le(x_1, x_2)) = 1 POL(s(x_1)) = 1 + x_1 POL(true) = 1 The following pairs are in P_>: IF(true, x, x_removed) -> D(s(x), x_removed) The following pairs are in P_bound: IF(true, x, x_removed) -> D(s(x), x_removed) The following rules are usable: true -> le(0, y) false -> le(s(x), 0) le(x, y) -> le(s(x), s(y)) ---------------------------------------- (133) Obligation: Q DP problem: The TRS P consists of the following rules: D(x, x_removed) -> IF(le(x, x_removed), x, x_removed) 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)) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) ack(0, x) -> s(x) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) ack(0, x0) ack(s(x0), 0) ack(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (134) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (135) TRUE