/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) TransformationProof [EQUIVALENT, 0 ms] (12) QDP (13) MNOCProof [EQUIVALENT, 0 ms] (14) QDP (15) NonTerminationLoopProof [COMPLETE, 0 ms] (16) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: g(f(x, y)) -> f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(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: g(f(x, y)) -> f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y)))) The set Q consists of the following terms: g(f(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: G(f(x, y)) -> G(g(x)) G(f(x, y)) -> G(x) G(f(x, y)) -> G(g(y)) G(f(x, y)) -> G(y) The TRS R consists of the following rules: g(f(x, y)) -> f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y)))) The set Q consists of the following terms: g(f(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule G(f(x, y)) -> G(g(x)) at position [0] we obtained the following new rules [LPAR04]: (G(f(f(x0, x1), y1)) -> G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1))))),G(f(f(x0, x1), y1)) -> G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))) ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: G(f(x, y)) -> G(x) G(f(x, y)) -> G(g(y)) G(f(x, y)) -> G(y) G(f(f(x0, x1), y1)) -> G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1))))) The TRS R consists of the following rules: g(f(x, y)) -> f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y)))) The set Q consists of the following terms: g(f(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule G(f(x, y)) -> G(g(y)) at position [0] we obtained the following new rules [LPAR04]: (G(f(y0, f(x0, x1))) -> G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1))))),G(f(y0, f(x0, x1))) -> G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))) ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: G(f(x, y)) -> G(x) G(f(x, y)) -> G(y) G(f(f(x0, x1), y1)) -> G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1))))) G(f(y0, f(x0, x1))) -> G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1))))) The TRS R consists of the following rules: g(f(x, y)) -> f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y)))) The set Q consists of the following terms: g(f(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule G(f(x, y)) -> G(x) we obtained the following new rules [LPAR04]: (G(f(f(y_0, y_1), x1)) -> G(f(y_0, y_1)),G(f(f(y_0, y_1), x1)) -> G(f(y_0, y_1))) (G(f(f(f(y_0, y_1), y_2), x1)) -> G(f(f(y_0, y_1), y_2)),G(f(f(f(y_0, y_1), y_2), x1)) -> G(f(f(y_0, y_1), y_2))) (G(f(f(y_0, f(y_1, y_2)), x1)) -> G(f(y_0, f(y_1, y_2))),G(f(f(y_0, f(y_1, y_2)), x1)) -> G(f(y_0, f(y_1, y_2)))) ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: G(f(x, y)) -> G(y) G(f(f(x0, x1), y1)) -> G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1))))) G(f(y0, f(x0, x1))) -> G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1))))) G(f(f(y_0, y_1), x1)) -> G(f(y_0, y_1)) G(f(f(f(y_0, y_1), y_2), x1)) -> G(f(f(y_0, y_1), y_2)) G(f(f(y_0, f(y_1, y_2)), x1)) -> G(f(y_0, f(y_1, y_2))) The TRS R consists of the following rules: g(f(x, y)) -> f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y)))) The set Q consists of the following terms: g(f(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule G(f(x, y)) -> G(y) we obtained the following new rules [LPAR04]: (G(f(x0, f(y_0, y_1))) -> G(f(y_0, y_1)),G(f(x0, f(y_0, y_1))) -> G(f(y_0, y_1))) (G(f(x0, f(f(y_0, y_1), y_2))) -> G(f(f(y_0, y_1), y_2)),G(f(x0, f(f(y_0, y_1), y_2))) -> G(f(f(y_0, y_1), y_2))) (G(f(x0, f(y_0, f(y_1, y_2)))) -> G(f(y_0, f(y_1, y_2))),G(f(x0, f(y_0, f(y_1, y_2)))) -> G(f(y_0, f(y_1, y_2)))) (G(f(x0, f(f(f(y_0, y_1), y_2), y_3))) -> G(f(f(f(y_0, y_1), y_2), y_3)),G(f(x0, f(f(f(y_0, y_1), y_2), y_3))) -> G(f(f(f(y_0, y_1), y_2), y_3))) (G(f(x0, f(f(y_0, f(y_1, y_2)), y_3))) -> G(f(f(y_0, f(y_1, y_2)), y_3)),G(f(x0, f(f(y_0, f(y_1, y_2)), y_3))) -> G(f(f(y_0, f(y_1, y_2)), y_3))) ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: G(f(f(x0, x1), y1)) -> G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1))))) G(f(y0, f(x0, x1))) -> G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1))))) G(f(f(y_0, y_1), x1)) -> G(f(y_0, y_1)) G(f(f(f(y_0, y_1), y_2), x1)) -> G(f(f(y_0, y_1), y_2)) G(f(f(y_0, f(y_1, y_2)), x1)) -> G(f(y_0, f(y_1, y_2))) G(f(x0, f(y_0, y_1))) -> G(f(y_0, y_1)) G(f(x0, f(f(y_0, y_1), y_2))) -> G(f(f(y_0, y_1), y_2)) G(f(x0, f(y_0, f(y_1, y_2)))) -> G(f(y_0, f(y_1, y_2))) G(f(x0, f(f(f(y_0, y_1), y_2), y_3))) -> G(f(f(f(y_0, y_1), y_2), y_3)) G(f(x0, f(f(y_0, f(y_1, y_2)), y_3))) -> G(f(f(y_0, f(y_1, y_2)), y_3)) The TRS R consists of the following rules: g(f(x, y)) -> f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y)))) The set Q consists of the following terms: g(f(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) MNOCProof (EQUIVALENT) We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: G(f(f(x0, x1), y1)) -> G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1))))) G(f(y0, f(x0, x1))) -> G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1))))) G(f(f(y_0, y_1), x1)) -> G(f(y_0, y_1)) G(f(f(f(y_0, y_1), y_2), x1)) -> G(f(f(y_0, y_1), y_2)) G(f(f(y_0, f(y_1, y_2)), x1)) -> G(f(y_0, f(y_1, y_2))) G(f(x0, f(y_0, y_1))) -> G(f(y_0, y_1)) G(f(x0, f(f(y_0, y_1), y_2))) -> G(f(f(y_0, y_1), y_2)) G(f(x0, f(y_0, f(y_1, y_2)))) -> G(f(y_0, f(y_1, y_2))) G(f(x0, f(f(f(y_0, y_1), y_2), y_3))) -> G(f(f(f(y_0, y_1), y_2), y_3)) G(f(x0, f(f(y_0, f(y_1, y_2)), y_3))) -> G(f(f(y_0, f(y_1, y_2)), y_3)) The TRS R consists of the following rules: g(f(x, y)) -> f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y)))) Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (15) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by semiunifying a rule from P directly. s = G(f(f(x0, x1), y1)) evaluates to t =G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1))))) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [x0 / g(g(x0)), x1 / g(g(x1)), y1 / f(g(g(x0)), g(g(x1)))] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from G(f(f(x0, x1), y1)) to G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1))))). ---------------------------------------- (16) NO