/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) MNOCProof [EQUIVALENT, 0 ms] (10) QDP (11) NonTerminationLoopProof [COMPLETE, 10 ms] (12) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(h(x), y) -> h(f(y, f(x, h(f(a, a))))) 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(h(x), y) -> h(f(y, f(x, h(f(a, a))))) The set Q consists of the following terms: f(h(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(h(x), y) -> F(y, f(x, h(f(a, a)))) F(h(x), y) -> F(x, h(f(a, a))) F(h(x), y) -> F(a, a) The TRS R consists of the following rules: f(h(x), y) -> h(f(y, f(x, h(f(a, a))))) The set Q consists of the following terms: f(h(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(h(x), y) -> F(x, h(f(a, a))) F(h(x), y) -> F(y, f(x, h(f(a, a)))) The TRS R consists of the following rules: f(h(x), y) -> h(f(y, f(x, h(f(a, a))))) The set Q consists of the following terms: f(h(x0), x1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule F(h(x), y) -> F(x, h(f(a, a))) we obtained the following new rules [LPAR04]: (F(h(h(y_0)), x1) -> F(h(y_0), h(f(a, a))),F(h(h(y_0)), x1) -> F(h(y_0), h(f(a, a)))) ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: F(h(x), y) -> F(y, f(x, h(f(a, a)))) F(h(h(y_0)), x1) -> F(h(y_0), h(f(a, a))) The TRS R consists of the following rules: f(h(x), y) -> h(f(y, f(x, h(f(a, a))))) The set Q consists of the following terms: f(h(x0), x1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) MNOCProof (EQUIVALENT) We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set. ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: F(h(x), y) -> F(y, f(x, h(f(a, a)))) F(h(h(y_0)), x1) -> F(h(y_0), h(f(a, a))) The TRS R consists of the following rules: f(h(x), y) -> h(f(y, f(x, h(f(a, a))))) Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (11) 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 = F(f(f(h(x), y''), y'), y) evaluates to t =F(y, f(f(y', f(f(y'', f(x, h(f(a, a)))), h(f(a, a)))), h(f(a, a)))) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [y'' / y''', x / x', x' / x'', x'' / x''', y''' / f(f(y'', f(x, h(f(a, a)))), h(f(a, a))), x''' / f(a, a)] * Semiunifier: [y' / h(x''), y / f(f(h(x'), y'''), h(x'''))] -------------------------------------------------------------------------------- Rewriting sequence F(f(f(h(x), y''), h(x'')), f(f(h(x'), y'''), h(x'''))) -> F(f(h(f(y'', f(x, h(f(a, a))))), h(x'')), f(f(h(x'), y'''), h(x'''))) with rule f(h(x1), y'''') -> h(f(y'''', f(x1, h(f(a, a))))) at position [0,0] and matcher [x1 / x, y'''' / y''] F(f(h(f(y'', f(x, h(f(a, a))))), h(x'')), f(f(h(x'), y'''), h(x'''))) -> F(h(f(h(x''), f(f(y'', f(x, h(f(a, a)))), h(f(a, a))))), f(f(h(x'), y'''), h(x'''))) with rule f(h(x'1), y') -> h(f(y', f(x'1, h(f(a, a))))) at position [0] and matcher [x'1 / f(y'', f(x, h(f(a, a)))), y' / h(x'')] F(h(f(h(x''), f(f(y'', f(x, h(f(a, a)))), h(f(a, a))))), f(f(h(x'), y'''), h(x'''))) -> F(f(f(h(x'), y'''), h(x''')), f(f(h(x''), f(f(y'', f(x, h(f(a, a)))), h(f(a, a)))), h(f(a, a)))) with rule F(h(x), y) -> F(y, f(x, h(f(a, a)))) 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. ---------------------------------------- (12) NO