/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- WORST_CASE(NON_POLY, ?) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF). (0) CpxTRS (1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] (2) TRS for Loop Detection (3) InfiniteLowerBoundProof [FINISHED, 0 ms] (4) BOUNDS(INF, INF) ---------------------------------------- (0) Obligation: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: from(X) -> cons(X, from(s(X))) head(cons(X, XS)) -> X 2nd(cons(X, XS)) -> head(XS) take(0, XS) -> nil take(s(N), cons(X, XS)) -> cons(X, take(N, XS)) sel(0, cons(X, XS)) -> X sel(s(N), cons(X, XS)) -> sel(N, XS) S is empty. Rewrite Strategy: FULL ---------------------------------------- (1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID)) Transformed a relative TRS into a decreasing-loop problem. ---------------------------------------- (2) Obligation: Analyzing the following TRS for decreasing loops: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: from(X) -> cons(X, from(s(X))) head(cons(X, XS)) -> X 2nd(cons(X, XS)) -> head(XS) take(0, XS) -> nil take(s(N), cons(X, XS)) -> cons(X, take(N, XS)) sel(0, cons(X, XS)) -> X sel(s(N), cons(X, XS)) -> sel(N, XS) S is empty. Rewrite Strategy: FULL ---------------------------------------- (3) InfiniteLowerBoundProof (FINISHED) The following loop proves infinite runtime complexity: The rewrite sequence from(X) ->^+ cons(X, from(s(X))) gives rise to a decreasing loop by considering the right hand sides subterm at position [1]. The pumping substitution is [ ]. The result substitution is [X / s(X)]. ---------------------------------------- (4) BOUNDS(INF, INF)