/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).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: msort(nil) -> nil msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y)))) min(x, nil) -> x min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z)) del(x, nil) -> nil del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z))) 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: msort(nil) -> nil msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y)))) min(x, nil) -> x min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z)) del(x, nil) -> nil del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z))) 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 del(x, .(y, z)) ->^+ if(=(x, y), z, .(y, del(x, z))) gives rise to a decreasing loop by considering the right hand sides subterm at position [2,1]. The pumping substitution is [z / .(y, z)]. The result substitution is [ ]. ---------------------------------------- (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: msort(nil) -> nil msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y)))) min(x, nil) -> x min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z)) del(x, nil) -> nil del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z))) 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: msort(nil) -> nil msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y)))) min(x, nil) -> x min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z)) del(x, nil) -> nil del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z))) S is empty. Rewrite Strategy: FULL ---------------------------------------- (9) DecreasingLoopProof (FINISHED) The following loop(s) give(s) rise to the lower bound EXP: The rewrite sequence min(x, .(y, z)) ->^+ if(<=(x, y), min(x, z), min(y, z)) gives rise to a decreasing loop by considering the right hand sides subterm at position [1]. The pumping substitution is [z / .(y, z)]. The result substitution is [ ]. The rewrite sequence min(x, .(y, z)) ->^+ if(<=(x, y), min(x, z), min(y, z)) gives rise to a decreasing loop by considering the right hand sides subterm at position [2]. The pumping substitution is [z / .(y, z)]. The result substitution is [x / y]. ---------------------------------------- (10) BOUNDS(EXP, INF)