/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 82 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 2 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 30 ms] (6) QDP (7) MRRProof [EQUIVALENT, 51 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) AND (11) QDP (12) UsableRulesProof [EQUIVALENT, 0 ms] (13) QDP (14) QDPSizeChangeProof [EQUIVALENT, 0 ms] (15) YES (16) QDP (17) MRRProof [EQUIVALENT, 6 ms] (18) QDP (19) MNOCProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPOrderProof [EQUIVALENT, 0 ms] (22) QDP (23) PisEmptyProof [EQUIVALENT, 0 ms] (24) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: P(x1) -> Q(Q(p(x1))) p(p(x1)) -> q(q(x1)) p(Q(Q(x1))) -> Q(Q(p(x1))) Q(p(q(x1))) -> q(p(Q(x1))) q(q(p(x1))) -> p(q(q(x1))) q(Q(x1)) -> x1 Q(q(x1)) -> x1 p(P(x1)) -> x1 P(p(x1)) -> x1 Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(P(x_1)) = 2 + x_1 POL(Q(x_1)) = 1 + x_1 POL(p(x_1)) = x_1 POL(q(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: q(Q(x1)) -> x1 Q(q(x1)) -> x1 p(P(x1)) -> x1 P(p(x1)) -> x1 ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: P(x1) -> Q(Q(p(x1))) p(p(x1)) -> q(q(x1)) p(Q(Q(x1))) -> Q(Q(p(x1))) Q(p(q(x1))) -> q(p(Q(x1))) q(q(p(x1))) -> p(q(q(x1))) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(P(x_1)) = 2 + x_1 POL(Q(x_1)) = x_1 POL(p(x_1)) = 1 + x_1 POL(q(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: P(x1) -> Q(Q(p(x1))) p(p(x1)) -> q(q(x1)) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(Q(Q(x1))) -> Q(Q(p(x1))) Q(p(q(x1))) -> q(p(Q(x1))) q(q(p(x1))) -> p(q(q(x1))) Q is empty. ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: P(Q(Q(x1))) -> Q^1(Q(p(x1))) P(Q(Q(x1))) -> Q^1(p(x1)) P(Q(Q(x1))) -> P(x1) Q^1(p(q(x1))) -> Q^2(p(Q(x1))) Q^1(p(q(x1))) -> P(Q(x1)) Q^1(p(q(x1))) -> Q^1(x1) Q^2(q(p(x1))) -> P(q(q(x1))) Q^2(q(p(x1))) -> Q^2(q(x1)) Q^2(q(p(x1))) -> Q^2(x1) The TRS R consists of the following rules: p(Q(Q(x1))) -> Q(Q(p(x1))) Q(p(q(x1))) -> q(p(Q(x1))) q(q(p(x1))) -> p(q(q(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) 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: Q^1(p(q(x1))) -> Q^2(p(Q(x1))) Q^1(p(q(x1))) -> P(Q(x1)) Q^1(p(q(x1))) -> Q^1(x1) Q^2(q(p(x1))) -> Q^2(x1) Strictly oriented rules of the TRS R: Q(p(q(x1))) -> q(p(Q(x1))) Used ordering: Polynomial interpretation [POLO]: POL(P(x_1)) = x_1 POL(Q(x_1)) = 2*x_1 POL(Q^1(x_1)) = 2*x_1 POL(Q^2(x_1)) = 1 + 2*x_1 POL(p(x_1)) = x_1 POL(q(x_1)) = 1 + 2*x_1 ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: P(Q(Q(x1))) -> Q^1(Q(p(x1))) P(Q(Q(x1))) -> Q^1(p(x1)) P(Q(Q(x1))) -> P(x1) Q^2(q(p(x1))) -> P(q(q(x1))) Q^2(q(p(x1))) -> Q^2(q(x1)) The TRS R consists of the following rules: p(Q(Q(x1))) -> Q(Q(p(x1))) q(q(p(x1))) -> p(q(q(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes. ---------------------------------------- (10) Complex Obligation (AND) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: P(Q(Q(x1))) -> P(x1) The TRS R consists of the following rules: p(Q(Q(x1))) -> Q(Q(p(x1))) q(q(p(x1))) -> p(q(q(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: P(Q(Q(x1))) -> P(x1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) 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: *P(Q(Q(x1))) -> P(x1) The graph contains the following edges 1 > 1 ---------------------------------------- (15) YES ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: Q^2(q(p(x1))) -> Q^2(q(x1)) The TRS R consists of the following rules: p(Q(Q(x1))) -> Q(Q(p(x1))) q(q(p(x1))) -> p(q(q(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) 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 rules of the TRS R: p(Q(Q(x1))) -> Q(Q(p(x1))) Used ordering: Polynomial interpretation [POLO]: POL(Q(x_1)) = 2 + 2*x_1 POL(Q^2(x_1)) = x_1 POL(p(x_1)) = 2*x_1 POL(q(x_1)) = 2*x_1 ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: Q^2(q(p(x1))) -> Q^2(q(x1)) The TRS R consists of the following rules: q(q(p(x1))) -> p(q(q(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: Q^2(q(p(x1))) -> Q^2(q(x1)) The TRS R consists of the following rules: q(q(p(x1))) -> p(q(q(x1))) The set Q consists of the following terms: q(q(p(x0))) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. Q^2(q(p(x1))) -> Q^2(q(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(Q^2(x_1)) = x_1 POL(p(x_1)) = 1 + x_1 POL(q(x_1)) = x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: q(q(p(x1))) -> p(q(q(x1))) ---------------------------------------- (22) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: q(q(p(x1))) -> p(q(q(x1))) The set Q consists of the following terms: q(q(p(x0))) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (24) YES