/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 61 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) MNOCProof [EQUIVALENT, 0 ms] (9) QDP (10) UsableRulesProof [EQUIVALENT, 0 ms] (11) QDP (12) QReductionProof [EQUIVALENT, 0 ms] (13) QDP (14) QDPSizeChangeProof [EQUIVALENT, 0 ms] (15) YES (16) QDP (17) MRRProof [EQUIVALENT, 13 ms] (18) QDP (19) PisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(a(a(x1)))) -> a(b(a(a(x1)))) a(b(a(a(x1)))) -> a(b(b(a(x1)))) b(b(b(a(x1)))) -> b(b(a(b(x1)))) b(a(b(a(x1)))) -> b(a(b(b(x1)))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a(x_1)) = 1 + x_1 POL(b(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a(b(a(a(x1)))) -> a(b(b(a(x1)))) b(a(b(a(x1)))) -> b(a(b(b(x1)))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(a(a(x1)))) -> a(b(a(a(x1)))) b(b(b(a(x1)))) -> b(b(a(b(x1)))) Q is empty. ---------------------------------------- (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: B(a(a(a(x1)))) -> B(a(a(x1))) B(b(b(a(x1)))) -> B(b(a(b(x1)))) B(b(b(a(x1)))) -> B(a(b(x1))) B(b(b(a(x1)))) -> B(x1) The TRS R consists of the following rules: b(a(a(a(x1)))) -> a(b(a(a(x1)))) b(b(b(a(x1)))) -> b(b(a(b(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: B(a(a(a(x1)))) -> B(a(a(x1))) The TRS R consists of the following rules: b(a(a(a(x1)))) -> a(b(a(a(x1)))) b(b(b(a(x1)))) -> b(b(a(b(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: B(a(a(a(x1)))) -> B(a(a(x1))) The TRS R consists of the following rules: b(a(a(a(x1)))) -> a(b(a(a(x1)))) b(b(b(a(x1)))) -> b(b(a(b(x1)))) The set Q consists of the following terms: b(a(a(a(x0)))) b(b(b(a(x0)))) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) 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. ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: B(a(a(a(x1)))) -> B(a(a(x1))) R is empty. The set Q consists of the following terms: b(a(a(a(x0)))) b(b(b(a(x0)))) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) 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]. b(a(a(a(x0)))) b(b(b(a(x0)))) ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: B(a(a(a(x1)))) -> B(a(a(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: *B(a(a(a(x1)))) -> B(a(a(x1))) The graph contains the following edges 1 > 1 ---------------------------------------- (15) YES ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: B(b(b(a(x1)))) -> B(x1) B(b(b(a(x1)))) -> B(b(a(b(x1)))) The TRS R consists of the following rules: b(a(a(a(x1)))) -> a(b(a(a(x1)))) b(b(b(a(x1)))) -> b(b(a(b(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 dependency pairs: B(b(b(a(x1)))) -> B(x1) B(b(b(a(x1)))) -> B(b(a(b(x1)))) Strictly oriented rules of the TRS R: b(a(a(a(x1)))) -> a(b(a(a(x1)))) b(b(b(a(x1)))) -> b(b(a(b(x1)))) Used ordering: Polynomial interpretation [POLO]: POL(B(x_1)) = 2*x_1 POL(a(x_1)) = 2 + 2*x_1 POL(b(x_1)) = 1 + 3*x_1 ---------------------------------------- (18) Obligation: Q DP problem: P is empty. R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (20) YES