/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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, 17 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, y, z) -> g(<=(x, y), x, y, z) g(true, x, y, z) -> z g(false, x, y, z) -> f(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)) p(0) -> 0 p(s(x)) -> x 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(x, y, z) -> g(<=(x, y), x, y, z) g(true, x, y, z) -> z g(false, x, y, z) -> f(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)) p(0) -> 0 p(s(x)) -> x The set Q consists of the following terms: f(x0, x1, x2) g(true, x0, x1, x2) g(false, x0, x1, x2) p(0) p(s(x0)) ---------------------------------------- (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(x, y, z) -> G(<=(x, y), x, y, z) G(false, x, y, z) -> F(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)) G(false, x, y, z) -> F(p(x), y, z) G(false, x, y, z) -> P(x) G(false, x, y, z) -> F(p(y), z, x) G(false, x, y, z) -> P(y) G(false, x, y, z) -> F(p(z), x, y) G(false, x, y, z) -> P(z) The TRS R consists of the following rules: f(x, y, z) -> g(<=(x, y), x, y, z) g(true, x, y, z) -> z g(false, x, y, z) -> f(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)) p(0) -> 0 p(s(x)) -> x The set Q consists of the following terms: f(x0, x1, x2) g(true, x0, x1, x2) g(false, x0, x1, x2) p(0) p(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 8 less nodes. ---------------------------------------- (6) TRUE