WORST_CASE(Omega(n^1),?) proof of /export/starexec/sandbox/benchmark/theBenchmark.trs # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, INF). (0) CpxRelTRS (1) SInnermostTerminationProof [BOTH CONCRETE BOUNDS(ID, ID), 6421 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 ---------------------------------------- (0) Obligation: The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, INF). The TRS R consists of the following rules: QSORT(nil) -> c QSORT(cons(z0, z1)) -> c1(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), QSORT(filterlow(last(cons(z0, z1)), cons(z0, z1))), FILTERLOW(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1))) QSORT(cons(z0, z1)) -> c2(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), LAST(cons(z0, z1))) QSORT(cons(z0, z1)) -> c3(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), QSORT(filterhigh(last(cons(z0, z1)), cons(z0, z1))), FILTERHIGH(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1))) FILTERLOW(z0, nil) -> c4 FILTERLOW(z0, cons(z1, z2)) -> c5(IF1(ge(z0, z1), z0, z1, z2), GE(z0, z1)) IF1(true, z0, z1, z2) -> c6(FILTERLOW(z0, z2)) IF1(false, z0, z1, z2) -> c7(FILTERLOW(z0, z2)) FILTERHIGH(z0, nil) -> c8 FILTERHIGH(z0, cons(z1, z2)) -> c9(IF2(ge(z1, z0), z0, z1, z2), GE(z1, z0)) IF2(true, z0, z1, z2) -> c10(FILTERHIGH(z0, z2)) IF2(false, z0, z1, z2) -> c11(FILTERHIGH(z0, z2)) GE(z0, 0) -> c12 GE(0, s(z0)) -> c13 GE(s(z0), s(z1)) -> c14(GE(z0, z1)) APPEND(nil, ys) -> c15 APPEND(cons(z0, z1), ys) -> c16(APPEND(z1, ys)) LAST(nil) -> c17 LAST(cons(z0, nil)) -> c18 LAST(cons(z0, cons(z1, z2))) -> c19(LAST(cons(z1, z2))) The (relative) TRS S consists of the following rules: qsort(nil) -> nil qsort(cons(z0, z1)) -> append(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))) filterlow(z0, nil) -> nil filterlow(z0, cons(z1, z2)) -> if1(ge(z0, z1), z0, z1, z2) if1(true, z0, z1, z2) -> filterlow(z0, z2) if1(false, z0, z1, z2) -> cons(z1, filterlow(z0, z2)) filterhigh(z0, nil) -> nil filterhigh(z0, cons(z1, z2)) -> if2(ge(z1, z0), z0, z1, z2) if2(true, z0, z1, z2) -> filterhigh(z0, z2) if2(false, z0, z1, z2) -> cons(z1, filterhigh(z0, z2)) ge(z0, 0) -> true ge(0, s(z0)) -> false ge(s(z0), s(z1)) -> ge(z0, z1) append(nil, ys) -> ys append(cons(z0, z1), ys) -> cons(z0, append(z1, ys)) last(nil) -> 0 last(cons(z0, nil)) -> z0 last(cons(z0, cons(z1, z2))) -> last(cons(z1, z2)) 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(n^1, INF). The TRS R consists of the following rules: QSORT(nil) -> c QSORT(cons(z0, z1)) -> c1(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), QSORT(filterlow(last(cons(z0, z1)), cons(z0, z1))), FILTERLOW(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1))) QSORT(cons(z0, z1)) -> c2(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), LAST(cons(z0, z1))) QSORT(cons(z0, z1)) -> c3(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), QSORT(filterhigh(last(cons(z0, z1)), cons(z0, z1))), FILTERHIGH(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1))) FILTERLOW(z0, nil) -> c4 FILTERLOW(z0, cons(z1, z2)) -> c5(IF1(ge(z0, z1), z0, z1, z2), GE(z0, z1)) IF1(true, z0, z1, z2) -> c6(FILTERLOW(z0, z2)) IF1(false, z0, z1, z2) -> c7(FILTERLOW(z0, z2)) FILTERHIGH(z0, nil) -> c8 FILTERHIGH(z0, cons(z1, z2)) -> c9(IF2(ge(z1, z0), z0, z1, z2), GE(z1, z0)) IF2(true, z0, z1, z2) -> c10(FILTERHIGH(z0, z2)) IF2(false, z0, z1, z2) -> c11(FILTERHIGH(z0, z2)) GE(z0, 0) -> c12 GE(0, s(z0)) -> c13 GE(s(z0), s(z1)) -> c14(GE(z0, z1)) APPEND(nil, ys) -> c15 APPEND(cons(z0, z1), ys) -> c16(APPEND(z1, ys)) LAST(nil) -> c17 LAST(cons(z0, nil)) -> c18 LAST(cons(z0, cons(z1, z2))) -> c19(LAST(cons(z1, z2))) The (relative) TRS S consists of the following rules: qsort(nil) -> nil qsort(cons(z0, z1)) -> append(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))) filterlow(z0, nil) -> nil filterlow(z0, cons(z1, z2)) -> if1(ge(z0, z1), z0, z1, z2) if1(true, z0, z1, z2) -> filterlow(z0, z2) if1(false, z0, z1, z2) -> cons(z1, filterlow(z0, z2)) filterhigh(z0, nil) -> nil filterhigh(z0, cons(z1, z2)) -> if2(ge(z1, z0), z0, z1, z2) if2(true, z0, z1, z2) -> filterhigh(z0, z2) if2(false, z0, z1, z2) -> cons(z1, filterhigh(z0, z2)) ge(z0, 0) -> true ge(0, s(z0)) -> false ge(s(z0), s(z1)) -> ge(z0, z1) append(nil, ys) -> ys append(cons(z0, z1), ys) -> cons(z0, append(z1, ys)) last(nil) -> 0 last(cons(z0, nil)) -> z0 last(cons(z0, cons(z1, z2))) -> last(cons(z1, z2)) 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(n^1, INF). The TRS R consists of the following rules: QSORT(nil) -> c QSORT(cons(z0, z1)) -> c1(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), QSORT(filterlow(last(cons(z0, z1)), cons(z0, z1))), FILTERLOW(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1))) QSORT(cons(z0, z1)) -> c2(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), LAST(cons(z0, z1))) QSORT(cons(z0, z1)) -> c3(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), QSORT(filterhigh(last(cons(z0, z1)), cons(z0, z1))), FILTERHIGH(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1))) FILTERLOW(z0, nil) -> c4 FILTERLOW(z0, cons(z1, z2)) -> c5(IF1(ge(z0, z1), z0, z1, z2), GE(z0, z1)) IF1(true, z0, z1, z2) -> c6(FILTERLOW(z0, z2)) IF1(false, z0, z1, z2) -> c7(FILTERLOW(z0, z2)) FILTERHIGH(z0, nil) -> c8 FILTERHIGH(z0, cons(z1, z2)) -> c9(IF2(ge(z1, z0), z0, z1, z2), GE(z1, z0)) IF2(true, z0, z1, z2) -> c10(FILTERHIGH(z0, z2)) IF2(false, z0, z1, z2) -> c11(FILTERHIGH(z0, z2)) GE(z0, 0) -> c12 GE(0, s(z0)) -> c13 GE(s(z0), s(z1)) -> c14(GE(z0, z1)) APPEND(nil, ys) -> c15 APPEND(cons(z0, z1), ys) -> c16(APPEND(z1, ys)) LAST(nil) -> c17 LAST(cons(z0, nil)) -> c18 LAST(cons(z0, cons(z1, z2))) -> c19(LAST(cons(z1, z2))) The (relative) TRS S consists of the following rules: qsort(nil) -> nil qsort(cons(z0, z1)) -> append(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))) filterlow(z0, nil) -> nil filterlow(z0, cons(z1, z2)) -> if1(ge(z0, z1), z0, z1, z2) if1(true, z0, z1, z2) -> filterlow(z0, z2) if1(false, z0, z1, z2) -> cons(z1, filterlow(z0, z2)) filterhigh(z0, nil) -> nil filterhigh(z0, cons(z1, z2)) -> if2(ge(z1, z0), z0, z1, z2) if2(true, z0, z1, z2) -> filterhigh(z0, z2) if2(false, z0, z1, z2) -> cons(z1, filterhigh(z0, z2)) ge(z0, 0) -> true ge(0, s(z0)) -> false ge(s(z0), s(z1)) -> ge(z0, z1) append(nil, ys) -> ys append(cons(z0, z1), ys) -> cons(z0, append(z1, ys)) last(nil) -> 0 last(cons(z0, nil)) -> z0 last(cons(z0, cons(z1, z2))) -> last(cons(z1, z2)) 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 APPEND(cons(z0, z1), ys) ->^+ c16(APPEND(z1, ys)) gives rise to a decreasing loop by considering the right hand sides subterm at position [0]. The pumping substitution is [z1 / cons(z0, z1)]. 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(n^1, INF). The TRS R consists of the following rules: QSORT(nil) -> c QSORT(cons(z0, z1)) -> c1(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), QSORT(filterlow(last(cons(z0, z1)), cons(z0, z1))), FILTERLOW(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1))) QSORT(cons(z0, z1)) -> c2(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), LAST(cons(z0, z1))) QSORT(cons(z0, z1)) -> c3(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), QSORT(filterhigh(last(cons(z0, z1)), cons(z0, z1))), FILTERHIGH(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1))) FILTERLOW(z0, nil) -> c4 FILTERLOW(z0, cons(z1, z2)) -> c5(IF1(ge(z0, z1), z0, z1, z2), GE(z0, z1)) IF1(true, z0, z1, z2) -> c6(FILTERLOW(z0, z2)) IF1(false, z0, z1, z2) -> c7(FILTERLOW(z0, z2)) FILTERHIGH(z0, nil) -> c8 FILTERHIGH(z0, cons(z1, z2)) -> c9(IF2(ge(z1, z0), z0, z1, z2), GE(z1, z0)) IF2(true, z0, z1, z2) -> c10(FILTERHIGH(z0, z2)) IF2(false, z0, z1, z2) -> c11(FILTERHIGH(z0, z2)) GE(z0, 0) -> c12 GE(0, s(z0)) -> c13 GE(s(z0), s(z1)) -> c14(GE(z0, z1)) APPEND(nil, ys) -> c15 APPEND(cons(z0, z1), ys) -> c16(APPEND(z1, ys)) LAST(nil) -> c17 LAST(cons(z0, nil)) -> c18 LAST(cons(z0, cons(z1, z2))) -> c19(LAST(cons(z1, z2))) The (relative) TRS S consists of the following rules: qsort(nil) -> nil qsort(cons(z0, z1)) -> append(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))) filterlow(z0, nil) -> nil filterlow(z0, cons(z1, z2)) -> if1(ge(z0, z1), z0, z1, z2) if1(true, z0, z1, z2) -> filterlow(z0, z2) if1(false, z0, z1, z2) -> cons(z1, filterlow(z0, z2)) filterhigh(z0, nil) -> nil filterhigh(z0, cons(z1, z2)) -> if2(ge(z1, z0), z0, z1, z2) if2(true, z0, z1, z2) -> filterhigh(z0, z2) if2(false, z0, z1, z2) -> cons(z1, filterhigh(z0, z2)) ge(z0, 0) -> true ge(0, s(z0)) -> false ge(s(z0), s(z1)) -> ge(z0, z1) append(nil, ys) -> ys append(cons(z0, z1), ys) -> cons(z0, append(z1, ys)) last(nil) -> 0 last(cons(z0, nil)) -> z0 last(cons(z0, cons(z1, z2))) -> last(cons(z1, z2)) 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(n^1, INF). The TRS R consists of the following rules: QSORT(nil) -> c QSORT(cons(z0, z1)) -> c1(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), QSORT(filterlow(last(cons(z0, z1)), cons(z0, z1))), FILTERLOW(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1))) QSORT(cons(z0, z1)) -> c2(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), LAST(cons(z0, z1))) QSORT(cons(z0, z1)) -> c3(APPEND(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))), QSORT(filterhigh(last(cons(z0, z1)), cons(z0, z1))), FILTERHIGH(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1))) FILTERLOW(z0, nil) -> c4 FILTERLOW(z0, cons(z1, z2)) -> c5(IF1(ge(z0, z1), z0, z1, z2), GE(z0, z1)) IF1(true, z0, z1, z2) -> c6(FILTERLOW(z0, z2)) IF1(false, z0, z1, z2) -> c7(FILTERLOW(z0, z2)) FILTERHIGH(z0, nil) -> c8 FILTERHIGH(z0, cons(z1, z2)) -> c9(IF2(ge(z1, z0), z0, z1, z2), GE(z1, z0)) IF2(true, z0, z1, z2) -> c10(FILTERHIGH(z0, z2)) IF2(false, z0, z1, z2) -> c11(FILTERHIGH(z0, z2)) GE(z0, 0) -> c12 GE(0, s(z0)) -> c13 GE(s(z0), s(z1)) -> c14(GE(z0, z1)) APPEND(nil, ys) -> c15 APPEND(cons(z0, z1), ys) -> c16(APPEND(z1, ys)) LAST(nil) -> c17 LAST(cons(z0, nil)) -> c18 LAST(cons(z0, cons(z1, z2))) -> c19(LAST(cons(z1, z2))) The (relative) TRS S consists of the following rules: qsort(nil) -> nil qsort(cons(z0, z1)) -> append(qsort(filterlow(last(cons(z0, z1)), cons(z0, z1))), cons(last(cons(z0, z1)), qsort(filterhigh(last(cons(z0, z1)), cons(z0, z1))))) filterlow(z0, nil) -> nil filterlow(z0, cons(z1, z2)) -> if1(ge(z0, z1), z0, z1, z2) if1(true, z0, z1, z2) -> filterlow(z0, z2) if1(false, z0, z1, z2) -> cons(z1, filterlow(z0, z2)) filterhigh(z0, nil) -> nil filterhigh(z0, cons(z1, z2)) -> if2(ge(z1, z0), z0, z1, z2) if2(true, z0, z1, z2) -> filterhigh(z0, z2) if2(false, z0, z1, z2) -> cons(z1, filterhigh(z0, z2)) ge(z0, 0) -> true ge(0, s(z0)) -> false ge(s(z0), s(z1)) -> ge(z0, z1) append(nil, ys) -> ys append(cons(z0, z1), ys) -> cons(z0, append(z1, ys)) last(nil) -> 0 last(cons(z0, nil)) -> z0 last(cons(z0, cons(z1, z2))) -> last(cons(z1, z2)) Rewrite Strategy: INNERMOST