/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). (0) CpxTRS (1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] (2) TRS for Loop Detection (3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms] (4) BEST (5) proven lower bound (6) LowerBoundPropagationProof [FINISHED, 0 ms] (7) BOUNDS(n^1, INF) (8) TRS for Loop Detection (9) DecreasingLoopProof [FINISHED, 0 ms] (10) BOUNDS(EXP, INF) ---------------------------------------- (0) Obligation: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). The TRS R consists of the following rules: rev(nil) -> nil rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l)) rev1(0, nil) -> 0 rev1(s(x), nil) -> s(x) rev1(x, cons(y, l)) -> rev1(y, l) rev2(x, nil) -> nil rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l))) 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(EXP, INF). The TRS R consists of the following rules: rev(nil) -> nil rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l)) rev1(0, nil) -> 0 rev1(s(x), nil) -> s(x) rev1(x, cons(y, l)) -> rev1(y, l) rev2(x, nil) -> nil rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l))) S is empty. Rewrite Strategy: FULL ---------------------------------------- (3) DecreasingLoopProof (LOWER BOUND(ID)) The following loop(s) give(s) rise to the lower bound Omega(n^1): The rewrite sequence rev2(x, cons(y, l)) ->^+ rev(cons(x, rev2(y, l))) gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1]. The pumping substitution is [l / cons(y, l)]. The result substitution is [x / y]. ---------------------------------------- (4) Complex Obligation (BEST) ---------------------------------------- (5) Obligation: Proved the lower bound n^1 for the following obligation: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). The TRS R consists of the following rules: rev(nil) -> nil rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l)) rev1(0, nil) -> 0 rev1(s(x), nil) -> s(x) rev1(x, cons(y, l)) -> rev1(y, l) rev2(x, nil) -> nil rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l))) S is empty. Rewrite Strategy: FULL ---------------------------------------- (6) LowerBoundPropagationProof (FINISHED) Propagated lower bound. ---------------------------------------- (7) BOUNDS(n^1, INF) ---------------------------------------- (8) Obligation: Analyzing the following TRS for decreasing loops: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). The TRS R consists of the following rules: rev(nil) -> nil rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l)) rev1(0, nil) -> 0 rev1(s(x), nil) -> s(x) rev1(x, cons(y, l)) -> rev1(y, l) rev2(x, nil) -> nil rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l))) S is empty. Rewrite Strategy: FULL ---------------------------------------- (9) DecreasingLoopProof (FINISHED) The following loop(s) give(s) rise to the lower bound EXP: The rewrite sequence rev2(x, cons(y, l)) ->^+ cons(rev1(x, rev2(y, l)), rev2(x, rev2(y, l))) gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1]. The pumping substitution is [l / cons(y, l)]. The result substitution is [x / y]. The rewrite sequence rev2(x, cons(y, l)) ->^+ cons(rev1(x, rev2(y, l)), rev2(x, rev2(y, l))) gives rise to a decreasing loop by considering the right hand sides subterm at position [1,1]. The pumping substitution is [l / cons(y, l)]. The result substitution is [x / y]. ---------------------------------------- (10) BOUNDS(EXP, INF)