/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QReductionProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 1 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0, y) -> 0 f(s(x), y) -> f(f(x, y), y) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0, y) -> 0 f(s(x), y) -> f(f(x, y), y) The set Q consists of the following terms: f(0, x0) f(s(x0), x1) ---------------------------------------- (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: F(s(x), y) -> F(f(x, y), y) F(s(x), y) -> F(x, y) The TRS R consists of the following rules: f(0, y) -> 0 f(s(x), y) -> f(f(x, y), y) The set Q consists of the following terms: f(0, x0) f(s(x0), x1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x), y) -> F(x, y) The TRS R consists of the following rules: f(0, y) -> 0 f(s(x), y) -> f(f(x, y), y) The set Q consists of the following terms: f(0, x0) f(s(x0), x1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) 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. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x), y) -> F(x, y) R is empty. The set Q consists of the following terms: f(0, x0) f(s(x0), x1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) 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(0, x0) f(s(x0), x1) ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x), y) -> F(x, y) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) 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: *F(s(x), y) -> F(x, y) The graph contains the following edges 1 > 1, 2 >= 2 ---------------------------------------- (12) YES