/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 25 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) NonTerminationLoopProof [COMPLETE, 0 ms] (8) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(f, 0), n) -> app(app(hd, app(app(map, f), app(app(cons, 0), nil))), n) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(f, 0), n) -> APP(app(hd, app(app(map, f), app(app(cons, 0), nil))), n) APP(app(f, 0), n) -> APP(hd, app(app(map, f), app(app(cons, 0), nil))) APP(app(f, 0), n) -> APP(app(map, f), app(app(cons, 0), nil)) APP(app(f, 0), n) -> APP(map, f) APP(app(f, 0), n) -> APP(app(cons, 0), nil) APP(app(f, 0), n) -> APP(cons, 0) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs)) APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x)) APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) The TRS R consists of the following rules: app(app(f, 0), n) -> app(app(hd, app(app(map, f), app(app(cons, 0), nil))), n) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(f, 0), n) -> APP(app(map, f), app(app(cons, 0), nil)) APP(app(f, 0), n) -> APP(app(hd, app(app(map, f), app(app(cons, 0), nil))), n) APP(app(f, 0), n) -> APP(app(cons, 0), nil) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs)) APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) The TRS R consists of the following rules: app(app(f, 0), n) -> app(app(hd, app(app(map, f), app(app(cons, 0), nil))), n) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule APP(app(f, 0), n) -> APP(app(hd, app(app(map, f), app(app(cons, 0), nil))), n) at position [0,1] we obtained the following new rules [LPAR04]: (APP(app(0, 0), y1) -> APP(app(hd, app(app(hd, app(app(map, map), app(app(cons, 0), nil))), app(app(cons, 0), nil))), y1),APP(app(0, 0), y1) -> APP(app(hd, app(app(hd, app(app(map, map), app(app(cons, 0), nil))), app(app(cons, 0), nil))), y1)) (APP(app(x0, 0), y1) -> APP(app(hd, app(app(cons, app(x0, 0)), app(app(map, x0), nil))), y1),APP(app(x0, 0), y1) -> APP(app(hd, app(app(cons, app(x0, 0)), app(app(map, x0), nil))), y1)) (APP(app(y0, 0), y1) -> APP(app(hd, app(app(map, y0), app(app(hd, app(app(map, cons), app(app(cons, 0), nil))), nil))), y1),APP(app(y0, 0), y1) -> APP(app(hd, app(app(map, y0), app(app(hd, app(app(map, cons), app(app(cons, 0), nil))), nil))), y1)) ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(f, 0), n) -> APP(app(map, f), app(app(cons, 0), nil)) APP(app(f, 0), n) -> APP(app(cons, 0), nil) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs)) APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) APP(app(0, 0), y1) -> APP(app(hd, app(app(hd, app(app(map, map), app(app(cons, 0), nil))), app(app(cons, 0), nil))), y1) APP(app(x0, 0), y1) -> APP(app(hd, app(app(cons, app(x0, 0)), app(app(map, x0), nil))), y1) APP(app(y0, 0), y1) -> APP(app(hd, app(app(map, y0), app(app(hd, app(app(map, cons), app(app(cons, 0), nil))), nil))), y1) The TRS R consists of the following rules: app(app(f, 0), n) -> app(app(hd, app(app(map, f), app(app(cons, 0), nil))), n) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) 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 = APP(app(f, 0), n) evaluates to t =APP(app(cons, 0), nil) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [f / cons, n / nil] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from APP(app(f, 0), n) to APP(app(cons, 0), nil). ---------------------------------------- (8) NO