/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 49 ms] (2) QTRS (3) Overlay + Local Confluence [EQUIVALENT, 0 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 23 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) UsableRulesProof [EQUIVALENT, 0 ms] (10) QDP (11) QReductionProof [EQUIVALENT, 0 ms] (12) QDP (13) MRRProof [EQUIVALENT, 10 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(x)) -> s(s(f(p(s(x))))) f(0) -> 0 p(s(x)) -> x Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(f(x_1)) = 1 + x_1 POL(p(x_1)) = x_1 POL(s(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: f(0) -> 0 ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(x)) -> s(s(f(p(s(x))))) p(s(x)) -> x Q is empty. ---------------------------------------- (3) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(x)) -> s(s(f(p(s(x))))) p(s(x)) -> x The set Q consists of the following terms: f(s(x0)) p(s(x0)) ---------------------------------------- (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: F(s(x)) -> F(p(s(x))) F(s(x)) -> P(s(x)) The TRS R consists of the following rules: f(s(x)) -> s(s(f(p(s(x))))) p(s(x)) -> x The set Q consists of the following terms: f(s(x0)) p(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x)) -> F(p(s(x))) The TRS R consists of the following rules: f(s(x)) -> s(s(f(p(s(x))))) p(s(x)) -> x The set Q consists of the following terms: f(s(x0)) p(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) 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. ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x)) -> F(p(s(x))) The TRS R consists of the following rules: p(s(x)) -> x The set Q consists of the following terms: f(s(x0)) p(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) 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]. f(s(x0)) ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x)) -> F(p(s(x))) The TRS R consists of the following rules: p(s(x)) -> x The set Q consists of the following terms: p(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) 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(s(x)) -> x Used ordering: Polynomial interpretation [POLO]: POL(F(x_1)) = x_1 POL(p(x_1)) = x_1 POL(s(x_1)) = 1 + x_1 ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x)) -> F(p(s(x))) R is empty. The set Q consists of the following terms: p(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (16) TRUE