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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) MRRProof [EQUIVALENT, 15 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) TransformationProof [EQUIVALENT, 0 ms] (12) QDP (13) SemLabProof [SOUND, 99 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) QDP (17) UsableRulesReductionPairsProof [EQUIVALENT, 6 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 0 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(p(b(a(x0)), x1), p(x2, x3)) -> p(p(b(x2), a(a(b(x1)))), p(x3, x0)) Q is empty. ---------------------------------------- (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: P(p(b(a(x0)), x1), p(x2, x3)) -> P(p(b(x2), a(a(b(x1)))), p(x3, x0)) P(p(b(a(x0)), x1), p(x2, x3)) -> P(b(x2), a(a(b(x1)))) P(p(b(a(x0)), x1), p(x2, x3)) -> P(x3, x0) The TRS R consists of the following rules: p(p(b(a(x0)), x1), p(x2, x3)) -> p(p(b(x2), a(a(b(x1)))), p(x3, x0)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(b(a(x0)), x1), p(x2, x3)) -> P(x3, x0) P(p(b(a(x0)), x1), p(x2, x3)) -> P(p(b(x2), a(a(b(x1)))), p(x3, x0)) The TRS R consists of the following rules: p(p(b(a(x0)), x1), p(x2, x3)) -> p(p(b(x2), a(a(b(x1)))), p(x3, x0)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: P(p(b(a(x0)), x1), p(x2, x3)) -> P(x3, x0) Used ordering: Polynomial interpretation [POLO]: POL(P(x_1, x_2)) = 2*x_1 + 2*x_2 POL(a(x_1)) = x_1 POL(b(x_1)) = x_1 POL(p(x_1, x_2)) = 1 + 2*x_1 + 2*x_2 ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(b(a(x0)), x1), p(x2, x3)) -> P(p(b(x2), a(a(b(x1)))), p(x3, x0)) The TRS R consists of the following rules: p(p(b(a(x0)), x1), p(x2, x3)) -> p(p(b(x2), a(a(b(x1)))), p(x3, x0)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule P(p(b(a(x0)), x1), p(x2, x3)) -> P(p(b(x2), a(a(b(x1)))), p(x3, x0)) we obtained the following new rules [LPAR04]: (P(p(b(a(x0)), a(a(b(y_1)))), p(x2, x3)) -> P(p(b(x2), a(a(b(a(a(b(y_1))))))), p(x3, x0)),P(p(b(a(x0)), a(a(b(y_1)))), p(x2, x3)) -> P(p(b(x2), a(a(b(a(a(b(y_1))))))), p(x3, x0))) ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(b(a(x0)), a(a(b(y_1)))), p(x2, x3)) -> P(p(b(x2), a(a(b(a(a(b(y_1))))))), p(x3, x0)) The TRS R consists of the following rules: p(p(b(a(x0)), x1), p(x2, x3)) -> p(p(b(x2), a(a(b(x1)))), p(x3, x0)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule P(p(b(a(x0)), a(a(b(y_1)))), p(x2, x3)) -> P(p(b(x2), a(a(b(a(a(b(y_1))))))), p(x3, x0)) we obtained the following new rules [LPAR04]: (P(p(b(a(x0)), a(a(b(a(a(b(y_1))))))), p(x2, x3)) -> P(p(b(x2), a(a(b(a(a(b(a(a(b(y_1)))))))))), p(x3, x0)),P(p(b(a(x0)), a(a(b(a(a(b(y_1))))))), p(x2, x3)) -> P(p(b(x2), a(a(b(a(a(b(a(a(b(y_1)))))))))), p(x3, x0))) ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(b(a(x0)), a(a(b(a(a(b(y_1))))))), p(x2, x3)) -> P(p(b(x2), a(a(b(a(a(b(a(a(b(y_1)))))))))), p(x3, x0)) The TRS R consists of the following rules: p(p(b(a(x0)), x1), p(x2, x3)) -> p(p(b(x2), a(a(b(x1)))), p(x3, x0)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule P(p(b(a(x0)), a(a(b(a(a(b(y_1))))))), p(x2, x3)) -> P(p(b(x2), a(a(b(a(a(b(a(a(b(y_1)))))))))), p(x3, x0)) we obtained the following new rules [LPAR04]: (P(p(b(a(x0)), a(a(b(a(a(b(x1))))))), p(a(y_0), x3)) -> P(p(b(a(y_0)), a(a(b(a(a(b(a(a(b(x1)))))))))), p(x3, x0)),P(p(b(a(x0)), a(a(b(a(a(b(x1))))))), p(a(y_0), x3)) -> P(p(b(a(y_0)), a(a(b(a(a(b(a(a(b(x1)))))))))), p(x3, x0))) ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(b(a(x0)), a(a(b(a(a(b(x1))))))), p(a(y_0), x3)) -> P(p(b(a(y_0)), a(a(b(a(a(b(a(a(b(x1)))))))))), p(x3, x0)) The TRS R consists of the following rules: p(p(b(a(x0)), x1), p(x2, x3)) -> p(p(b(x2), a(a(b(x1)))), p(x3, x0)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) SemLabProof (SOUND) We found the following model for the rules of the TRSs R and P. Interpretation over the domain with elements from 0 to 1. a: 1 p: 0 b: 0 P: 0 By semantic labelling [SEMLAB] we obtain the following labelled QDP problem. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: P.0-0(p.0-1(b.1(a.0(x0)), a.1(a.0(b.1(a.1(a.0(b.0(x1))))))), p.1-0(a.0(y_0), x3)) -> P.0-0(p.0-1(b.1(a.0(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.0(x1)))))))))), p.0-0(x3, x0)) P.0-0(p.0-1(b.1(a.0(x0)), a.1(a.0(b.1(a.1(a.0(b.0(x1))))))), p.1-1(a.0(y_0), x3)) -> P.0-0(p.0-1(b.1(a.0(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.0(x1)))))))))), p.1-0(x3, x0)) P.0-0(p.0-1(b.1(a.0(x0)), a.1(a.0(b.1(a.1(a.0(b.0(x1))))))), p.1-0(a.1(y_0), x3)) -> P.0-0(p.0-1(b.1(a.1(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.0(x1)))))))))), p.0-0(x3, x0)) P.0-0(p.0-1(b.1(a.0(x0)), a.1(a.0(b.1(a.1(a.0(b.0(x1))))))), p.1-1(a.1(y_0), x3)) -> P.0-0(p.0-1(b.1(a.1(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.0(x1)))))))))), p.1-0(x3, x0)) P.0-0(p.0-1(b.1(a.0(x0)), a.1(a.0(b.1(a.1(a.0(b.1(x1))))))), p.1-0(a.0(y_0), x3)) -> P.0-0(p.0-1(b.1(a.0(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.1(x1)))))))))), p.0-0(x3, x0)) P.0-0(p.0-1(b.1(a.0(x0)), a.1(a.0(b.1(a.1(a.0(b.1(x1))))))), p.1-1(a.0(y_0), x3)) -> P.0-0(p.0-1(b.1(a.0(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.1(x1)))))))))), p.1-0(x3, x0)) P.0-0(p.0-1(b.1(a.0(x0)), a.1(a.0(b.1(a.1(a.0(b.1(x1))))))), p.1-0(a.1(y_0), x3)) -> P.0-0(p.0-1(b.1(a.1(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.1(x1)))))))))), p.0-0(x3, x0)) P.0-0(p.0-1(b.1(a.0(x0)), a.1(a.0(b.1(a.1(a.0(b.1(x1))))))), p.1-1(a.1(y_0), x3)) -> P.0-0(p.0-1(b.1(a.1(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.1(x1)))))))))), p.1-0(x3, x0)) P.0-0(p.0-1(b.1(a.1(x0)), a.1(a.0(b.1(a.1(a.0(b.0(x1))))))), p.1-0(a.0(y_0), x3)) -> P.0-0(p.0-1(b.1(a.0(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.0(x1)))))))))), p.0-1(x3, x0)) P.0-0(p.0-1(b.1(a.1(x0)), a.1(a.0(b.1(a.1(a.0(b.0(x1))))))), p.1-1(a.0(y_0), x3)) -> P.0-0(p.0-1(b.1(a.0(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.0(x1)))))))))), p.1-1(x3, x0)) P.0-0(p.0-1(b.1(a.1(x0)), a.1(a.0(b.1(a.1(a.0(b.0(x1))))))), p.1-0(a.1(y_0), x3)) -> P.0-0(p.0-1(b.1(a.1(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.0(x1)))))))))), p.0-1(x3, x0)) P.0-0(p.0-1(b.1(a.1(x0)), a.1(a.0(b.1(a.1(a.0(b.0(x1))))))), p.1-1(a.1(y_0), x3)) -> P.0-0(p.0-1(b.1(a.1(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.0(x1)))))))))), p.1-1(x3, x0)) P.0-0(p.0-1(b.1(a.1(x0)), a.1(a.0(b.1(a.1(a.0(b.1(x1))))))), p.1-0(a.0(y_0), x3)) -> P.0-0(p.0-1(b.1(a.0(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.1(x1)))))))))), p.0-1(x3, x0)) P.0-0(p.0-1(b.1(a.1(x0)), a.1(a.0(b.1(a.1(a.0(b.1(x1))))))), p.1-1(a.0(y_0), x3)) -> P.0-0(p.0-1(b.1(a.0(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.1(x1)))))))))), p.1-1(x3, x0)) P.0-0(p.0-1(b.1(a.1(x0)), a.1(a.0(b.1(a.1(a.0(b.1(x1))))))), p.1-0(a.1(y_0), x3)) -> P.0-0(p.0-1(b.1(a.1(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.1(x1)))))))))), p.0-1(x3, x0)) P.0-0(p.0-1(b.1(a.1(x0)), a.1(a.0(b.1(a.1(a.0(b.1(x1))))))), p.1-1(a.1(y_0), x3)) -> P.0-0(p.0-1(b.1(a.1(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.1(x1)))))))))), p.1-1(x3, x0)) The TRS R consists of the following rules: p.0-0(p.0-0(b.1(a.0(x0)), x1), p.0-0(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.0(x1)))), p.0-0(x3, x0)) p.0-0(p.0-0(b.1(a.0(x0)), x1), p.0-1(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.0(x1)))), p.1-0(x3, x0)) p.0-0(p.0-0(b.1(a.0(x0)), x1), p.1-0(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.0(x1)))), p.0-0(x3, x0)) p.0-0(p.0-0(b.1(a.0(x0)), x1), p.1-1(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.0(x1)))), p.1-0(x3, x0)) p.0-0(p.0-1(b.1(a.0(x0)), x1), p.0-0(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.1(x1)))), p.0-0(x3, x0)) p.0-0(p.0-1(b.1(a.0(x0)), x1), p.0-1(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.1(x1)))), p.1-0(x3, x0)) p.0-0(p.0-1(b.1(a.0(x0)), x1), p.1-0(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.1(x1)))), p.0-0(x3, x0)) p.0-0(p.0-1(b.1(a.0(x0)), x1), p.1-1(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.1(x1)))), p.1-0(x3, x0)) p.0-0(p.0-0(b.1(a.1(x0)), x1), p.0-0(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.0(x1)))), p.0-1(x3, x0)) p.0-0(p.0-0(b.1(a.1(x0)), x1), p.0-1(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.0(x1)))), p.1-1(x3, x0)) p.0-0(p.0-0(b.1(a.1(x0)), x1), p.1-0(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.0(x1)))), p.0-1(x3, x0)) p.0-0(p.0-0(b.1(a.1(x0)), x1), p.1-1(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.0(x1)))), p.1-1(x3, x0)) p.0-0(p.0-1(b.1(a.1(x0)), x1), p.0-0(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.1(x1)))), p.0-1(x3, x0)) p.0-0(p.0-1(b.1(a.1(x0)), x1), p.0-1(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.1(x1)))), p.1-1(x3, x0)) p.0-0(p.0-1(b.1(a.1(x0)), x1), p.1-0(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.1(x1)))), p.0-1(x3, x0)) p.0-0(p.0-1(b.1(a.1(x0)), x1), p.1-1(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.1(x1)))), p.1-1(x3, x0)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 15 less nodes. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: P.0-0(p.0-1(b.1(a.1(x0)), a.1(a.0(b.1(a.1(a.0(b.1(x1))))))), p.1-1(a.1(y_0), x3)) -> P.0-0(p.0-1(b.1(a.1(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.1(x1)))))))))), p.1-1(x3, x0)) The TRS R consists of the following rules: p.0-0(p.0-0(b.1(a.0(x0)), x1), p.0-0(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.0(x1)))), p.0-0(x3, x0)) p.0-0(p.0-0(b.1(a.0(x0)), x1), p.0-1(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.0(x1)))), p.1-0(x3, x0)) p.0-0(p.0-0(b.1(a.0(x0)), x1), p.1-0(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.0(x1)))), p.0-0(x3, x0)) p.0-0(p.0-0(b.1(a.0(x0)), x1), p.1-1(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.0(x1)))), p.1-0(x3, x0)) p.0-0(p.0-1(b.1(a.0(x0)), x1), p.0-0(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.1(x1)))), p.0-0(x3, x0)) p.0-0(p.0-1(b.1(a.0(x0)), x1), p.0-1(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.1(x1)))), p.1-0(x3, x0)) p.0-0(p.0-1(b.1(a.0(x0)), x1), p.1-0(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.1(x1)))), p.0-0(x3, x0)) p.0-0(p.0-1(b.1(a.0(x0)), x1), p.1-1(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.1(x1)))), p.1-0(x3, x0)) p.0-0(p.0-0(b.1(a.1(x0)), x1), p.0-0(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.0(x1)))), p.0-1(x3, x0)) p.0-0(p.0-0(b.1(a.1(x0)), x1), p.0-1(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.0(x1)))), p.1-1(x3, x0)) p.0-0(p.0-0(b.1(a.1(x0)), x1), p.1-0(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.0(x1)))), p.0-1(x3, x0)) p.0-0(p.0-0(b.1(a.1(x0)), x1), p.1-1(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.0(x1)))), p.1-1(x3, x0)) p.0-0(p.0-1(b.1(a.1(x0)), x1), p.0-0(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.1(x1)))), p.0-1(x3, x0)) p.0-0(p.0-1(b.1(a.1(x0)), x1), p.0-1(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.1(x1)))), p.1-1(x3, x0)) p.0-0(p.0-1(b.1(a.1(x0)), x1), p.1-0(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.1(x1)))), p.0-1(x3, x0)) p.0-0(p.0-1(b.1(a.1(x0)), x1), p.1-1(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.1(x1)))), p.1-1(x3, x0)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) UsableRulesReductionPairsProof (EQUIVALENT) By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well. No dependency pairs are removed. The following rules are removed from R: p.0-0(p.0-0(b.1(a.0(x0)), x1), p.0-0(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.0(x1)))), p.0-0(x3, x0)) p.0-0(p.0-0(b.1(a.0(x0)), x1), p.0-1(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.0(x1)))), p.1-0(x3, x0)) p.0-0(p.0-0(b.1(a.0(x0)), x1), p.1-0(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.0(x1)))), p.0-0(x3, x0)) p.0-0(p.0-0(b.1(a.0(x0)), x1), p.1-1(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.0(x1)))), p.1-0(x3, x0)) p.0-0(p.0-1(b.1(a.0(x0)), x1), p.0-0(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.1(x1)))), p.0-0(x3, x0)) p.0-0(p.0-1(b.1(a.0(x0)), x1), p.0-1(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.1(x1)))), p.1-0(x3, x0)) p.0-0(p.0-1(b.1(a.0(x0)), x1), p.1-0(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.1(x1)))), p.0-0(x3, x0)) p.0-0(p.0-1(b.1(a.0(x0)), x1), p.1-1(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.1(x1)))), p.1-0(x3, x0)) p.0-0(p.0-0(b.1(a.1(x0)), x1), p.0-0(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.0(x1)))), p.0-1(x3, x0)) p.0-0(p.0-0(b.1(a.1(x0)), x1), p.0-1(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.0(x1)))), p.1-1(x3, x0)) p.0-0(p.0-0(b.1(a.1(x0)), x1), p.1-0(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.0(x1)))), p.0-1(x3, x0)) p.0-0(p.0-0(b.1(a.1(x0)), x1), p.1-1(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.0(x1)))), p.1-1(x3, x0)) p.0-0(p.0-1(b.1(a.1(x0)), x1), p.0-0(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.1(x1)))), p.0-1(x3, x0)) p.0-0(p.0-1(b.1(a.1(x0)), x1), p.0-1(x2, x3)) -> p.0-0(p.0-1(b.0(x2), a.1(a.0(b.1(x1)))), p.1-1(x3, x0)) p.0-0(p.0-1(b.1(a.1(x0)), x1), p.1-0(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.1(x1)))), p.0-1(x3, x0)) p.0-0(p.0-1(b.1(a.1(x0)), x1), p.1-1(x2, x3)) -> p.0-0(p.0-1(b.1(x2), a.1(a.0(b.1(x1)))), p.1-1(x3, x0)) Used ordering: POLO with Polynomial interpretation [POLO]: POL(P.0-0(x_1, x_2)) = x_1 + x_2 POL(a.0(x_1)) = x_1 POL(a.1(x_1)) = x_1 POL(b.1(x_1)) = x_1 POL(p.0-1(x_1, x_2)) = x_1 + x_2 POL(p.1-1(x_1, x_2)) = x_1 + x_2 ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: P.0-0(p.0-1(b.1(a.1(x0)), a.1(a.0(b.1(a.1(a.0(b.1(x1))))))), p.1-1(a.1(y_0), x3)) -> P.0-0(p.0-1(b.1(a.1(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.1(x1)))))))))), p.1-1(x3, x0)) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. P.0-0(p.0-1(b.1(a.1(x0)), a.1(a.0(b.1(a.1(a.0(b.1(x1))))))), p.1-1(a.1(y_0), x3)) -> P.0-0(p.0-1(b.1(a.1(y_0)), a.1(a.0(b.1(a.1(a.0(b.1(a.1(a.0(b.1(x1)))))))))), p.1-1(x3, x0)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(P.0-0(x_1, x_2)) = x_1 + x_2 POL(a.0(x_1)) = 1 POL(a.1(x_1)) = 1 + x_1 POL(b.1(x_1)) = 1 + x_1 POL(p.0-1(x_1, x_2)) = x_1 + x_2 POL(p.1-1(x_1, x_2)) = x_1 + x_2 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: none ---------------------------------------- (20) Obligation: Q DP problem: P is empty. R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (22) YES