/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 68 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 6 ms] (6) QTRS (7) DependencyPairsProof [EQUIVALENT, 0 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (12) QDP (13) NonTerminationLoopProof [COMPLETE, 10 ms] (14) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__f(b, X, c) -> a__f(X, a__c, X) a__c -> b mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) mark(c) -> a__c mark(b) -> b a__f(X1, X2, X3) -> f(X1, X2, X3) a__c -> c Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a__c) = 0 POL(a__f(x_1, x_2, x_3)) = 2 + x_1 + 2*x_2 + x_3 POL(b) = 0 POL(c) = 0 POL(f(x_1, x_2, x_3)) = 2 + x_1 + 2*x_2 + x_3 POL(mark(x_1)) = 2*x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__f(b, X, c) -> a__f(X, a__c, X) a__c -> b mark(c) -> a__c mark(b) -> b a__f(X1, X2, X3) -> f(X1, X2, X3) a__c -> c Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a__c) = 0 POL(a__f(x_1, x_2, x_3)) = 1 + x_1 + 2*x_2 + x_3 POL(b) = 0 POL(c) = 0 POL(f(x_1, x_2, x_3)) = x_1 + 2*x_2 + x_3 POL(mark(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__f(X1, X2, X3) -> f(X1, X2, X3) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__f(b, X, c) -> a__f(X, a__c, X) a__c -> b mark(c) -> a__c mark(b) -> b a__c -> c Q is empty. ---------------------------------------- (5) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a__c) = 0 POL(a__f(x_1, x_2, x_3)) = x_1 + 2*x_2 + x_3 POL(b) = 0 POL(c) = 0 POL(mark(x_1)) = 2 + 2*x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: mark(c) -> a__c mark(b) -> b ---------------------------------------- (6) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__f(b, X, c) -> a__f(X, a__c, X) a__c -> b a__c -> c Q is empty. ---------------------------------------- (7) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: A__F(b, X, c) -> A__F(X, a__c, X) A__F(b, X, c) -> A__C The TRS R consists of the following rules: a__f(b, X, c) -> a__f(X, a__c, X) a__c -> b a__c -> c 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 1 SCC with 1 less node. ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: A__F(b, X, c) -> A__F(X, a__c, X) The TRS R consists of the following rules: a__f(b, X, c) -> a__f(X, a__c, X) a__c -> b a__c -> c Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) 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: a__f(b, X, c) -> a__f(X, a__c, X) Used ordering: POLO with Polynomial interpretation [POLO]: POL(A__F(x_1, x_2, x_3)) = x_1 + 2*x_2 + x_3 POL(a__c) = 0 POL(b) = 0 POL(c) = 0 ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: A__F(b, X, c) -> A__F(X, a__c, X) The TRS R consists of the following rules: a__c -> b a__c -> c Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: s = A__F(a__c, X, a__c) evaluates to t =A__F(X, a__c, X) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [X / a__c] -------------------------------------------------------------------------------- Rewriting sequence A__F(a__c, a__c, a__c) -> A__F(a__c, a__c, c) with rule a__c -> c at position [2] and matcher [ ] A__F(a__c, a__c, c) -> A__F(b, a__c, c) with rule a__c -> b at position [0] and matcher [ ] A__F(b, a__c, c) -> A__F(a__c, a__c, a__c) with rule A__F(b, X, c) -> A__F(X, a__c, X) Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence All these steps are and every following step will be a correct step w.r.t to Q. ---------------------------------------- (14) NO