/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(NON_POLY, ?) proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(INF, INF). (0) CpxRelTRS (1) SInnermostTerminationProof [BOTH CONCRETE BOUNDS(ID, ID), 186 ms] (2) CpxRelTRS (3) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] (4) TRS for Loop Detection (5) DecreasingLoopProof [LOWER BOUND(ID), 0 ms] (6) BEST (7) proven lower bound (8) LowerBoundPropagationProof [FINISHED, 0 ms] (9) BOUNDS(n^1, INF) (10) TRS for Loop Detection (11) InfiniteLowerBoundProof [FINISHED, 425 ms] (12) BOUNDS(INF, INF) ---------------------------------------- (0) Obligation: The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: lt0(Cons(x', xs'), Cons(x, xs)) -> lt0(xs', xs) g(x, Nil) -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) f(x, Nil) -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) lt0(x, Nil) -> False g(x, Cons(x', xs)) -> g[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs)) f(x, Cons(x', xs)) -> f(f[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs)), f[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs))) number42 -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) goal(x, y) -> Cons(f(x, y), Cons(g(x, y), Nil)) The (relative) TRS S consists of the following rules: g[Ite][False][Ite](False, Cons(x, xs), y) -> g(xs, Cons(Cons(Nil, Nil), y)) g[Ite][False][Ite](True, x', Cons(x, xs)) -> g(x', xs) f[Ite][False][Ite](False, Cons(x, xs), y) -> xs f[Ite][False][Ite](True, x', Cons(x, xs)) -> xs f[Ite][False][Ite](False, x, y) -> Cons(Cons(Nil, Nil), y) f[Ite][False][Ite](True, x, y) -> x Rewrite Strategy: INNERMOST ---------------------------------------- (1) SInnermostTerminationProof (BOTH CONCRETE BOUNDS(ID, ID)) proved innermost termination of relative rules ---------------------------------------- (2) Obligation: The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: lt0(Cons(x', xs'), Cons(x, xs)) -> lt0(xs', xs) g(x, Nil) -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) f(x, Nil) -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) lt0(x, Nil) -> False g(x, Cons(x', xs)) -> g[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs)) f(x, Cons(x', xs)) -> f(f[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs)), f[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs))) number42 -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) goal(x, y) -> Cons(f(x, y), Cons(g(x, y), Nil)) The (relative) TRS S consists of the following rules: g[Ite][False][Ite](False, Cons(x, xs), y) -> g(xs, Cons(Cons(Nil, Nil), y)) g[Ite][False][Ite](True, x', Cons(x, xs)) -> g(x', xs) f[Ite][False][Ite](False, Cons(x, xs), y) -> xs f[Ite][False][Ite](True, x', Cons(x, xs)) -> xs f[Ite][False][Ite](False, x, y) -> Cons(Cons(Nil, Nil), y) f[Ite][False][Ite](True, x, y) -> x Rewrite Strategy: INNERMOST ---------------------------------------- (3) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID)) Transformed a relative TRS into a decreasing-loop problem. ---------------------------------------- (4) Obligation: Analyzing the following TRS for decreasing loops: The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: lt0(Cons(x', xs'), Cons(x, xs)) -> lt0(xs', xs) g(x, Nil) -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) f(x, Nil) -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) lt0(x, Nil) -> False g(x, Cons(x', xs)) -> g[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs)) f(x, Cons(x', xs)) -> f(f[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs)), f[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs))) number42 -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) goal(x, y) -> Cons(f(x, y), Cons(g(x, y), Nil)) The (relative) TRS S consists of the following rules: g[Ite][False][Ite](False, Cons(x, xs), y) -> g(xs, Cons(Cons(Nil, Nil), y)) g[Ite][False][Ite](True, x', Cons(x, xs)) -> g(x', xs) f[Ite][False][Ite](False, Cons(x, xs), y) -> xs f[Ite][False][Ite](True, x', Cons(x, xs)) -> xs f[Ite][False][Ite](False, x, y) -> Cons(Cons(Nil, Nil), y) f[Ite][False][Ite](True, x, y) -> x Rewrite Strategy: INNERMOST ---------------------------------------- (5) DecreasingLoopProof (LOWER BOUND(ID)) The following loop(s) give(s) rise to the lower bound Omega(n^1): The rewrite sequence lt0(Cons(x', xs'), Cons(x, xs)) ->^+ lt0(xs', xs) gives rise to a decreasing loop by considering the right hand sides subterm at position []. The pumping substitution is [xs' / Cons(x', xs'), xs / Cons(x, xs)]. The result substitution is [ ]. ---------------------------------------- (6) Complex Obligation (BEST) ---------------------------------------- (7) Obligation: Proved the lower bound n^1 for the following obligation: The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: lt0(Cons(x', xs'), Cons(x, xs)) -> lt0(xs', xs) g(x, Nil) -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) f(x, Nil) -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) lt0(x, Nil) -> False g(x, Cons(x', xs)) -> g[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs)) f(x, Cons(x', xs)) -> f(f[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs)), f[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs))) number42 -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) goal(x, y) -> Cons(f(x, y), Cons(g(x, y), Nil)) The (relative) TRS S consists of the following rules: g[Ite][False][Ite](False, Cons(x, xs), y) -> g(xs, Cons(Cons(Nil, Nil), y)) g[Ite][False][Ite](True, x', Cons(x, xs)) -> g(x', xs) f[Ite][False][Ite](False, Cons(x, xs), y) -> xs f[Ite][False][Ite](True, x', Cons(x, xs)) -> xs f[Ite][False][Ite](False, x, y) -> Cons(Cons(Nil, Nil), y) f[Ite][False][Ite](True, x, y) -> x Rewrite Strategy: INNERMOST ---------------------------------------- (8) LowerBoundPropagationProof (FINISHED) Propagated lower bound. ---------------------------------------- (9) BOUNDS(n^1, INF) ---------------------------------------- (10) Obligation: Analyzing the following TRS for decreasing loops: The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: lt0(Cons(x', xs'), Cons(x, xs)) -> lt0(xs', xs) g(x, Nil) -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) f(x, Nil) -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) lt0(x, Nil) -> False g(x, Cons(x', xs)) -> g[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs)) f(x, Cons(x', xs)) -> f(f[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs)), f[Ite][False][Ite](lt0(x, Cons(Nil, Nil)), x, Cons(x', xs))) number42 -> Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))))))))))))))))))))))))))) goal(x, y) -> Cons(f(x, y), Cons(g(x, y), Nil)) The (relative) TRS S consists of the following rules: g[Ite][False][Ite](False, Cons(x, xs), y) -> g(xs, Cons(Cons(Nil, Nil), y)) g[Ite][False][Ite](True, x', Cons(x, xs)) -> g(x', xs) f[Ite][False][Ite](False, Cons(x, xs), y) -> xs f[Ite][False][Ite](True, x', Cons(x, xs)) -> xs f[Ite][False][Ite](False, x, y) -> Cons(Cons(Nil, Nil), y) f[Ite][False][Ite](True, x, y) -> x Rewrite Strategy: INNERMOST ---------------------------------------- (11) InfiniteLowerBoundProof (FINISHED) The following loop proves infinite runtime complexity: The rewrite sequence f(Cons(x'1_0, xs'2_0), Cons(x', xs)) ->^+ f(Cons(Cons(Nil, Nil), Cons(x', xs)), Cons(Cons(Nil, Nil), Cons(x', xs))) gives rise to a decreasing loop by considering the right hand sides subterm at position []. The pumping substitution is [ ]. The result substitution is [x'1_0 / Cons(Nil, Nil), xs'2_0 / Cons(x', xs), x' / Cons(Nil, Nil), xs / Cons(x', xs)]. ---------------------------------------- (12) BOUNDS(INF, INF)