/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).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, 63 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: dbl(0) -> 0 dbl(s(X)) -> s(n__s(n__dbl(activate(X)))) dbls(nil) -> nil dbls(cons(X, Y)) -> cons(n__dbl(activate(X)), n__dbls(activate(Y))) sel(0, cons(X, Y)) -> activate(X) sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z)) indx(nil, X) -> nil indx(cons(X, Y), Z) -> cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z))) from(X) -> cons(activate(X), n__from(n__s(activate(X)))) dbl1(0) -> 01 dbl1(s(X)) -> s1(s1(dbl1(activate(X)))) sel1(0, cons(X, Y)) -> activate(X) sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z)) quote(0) -> 01 quote(s(X)) -> s1(quote(activate(X))) quote(dbl(X)) -> dbl1(X) quote(sel(X, Y)) -> sel1(X, Y) s(X) -> n__s(X) dbl(X) -> n__dbl(X) dbls(X) -> n__dbls(X) sel(X1, X2) -> n__sel(X1, X2) indx(X1, X2) -> n__indx(X1, X2) from(X) -> n__from(X) activate(n__s(X)) -> s(X) activate(n__dbl(X)) -> dbl(activate(X)) activate(n__dbls(X)) -> dbls(activate(X)) activate(n__sel(X1, X2)) -> sel(activate(X1), activate(X2)) activate(n__indx(X1, X2)) -> indx(activate(X1), X2) activate(n__from(X)) -> from(X) activate(X) -> X 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: dbl(0) -> 0 dbl(s(X)) -> s(n__s(n__dbl(activate(X)))) dbls(nil) -> nil dbls(cons(X, Y)) -> cons(n__dbl(activate(X)), n__dbls(activate(Y))) sel(0, cons(X, Y)) -> activate(X) sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z)) indx(nil, X) -> nil indx(cons(X, Y), Z) -> cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z))) from(X) -> cons(activate(X), n__from(n__s(activate(X)))) dbl1(0) -> 01 dbl1(s(X)) -> s1(s1(dbl1(activate(X)))) sel1(0, cons(X, Y)) -> activate(X) sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z)) quote(0) -> 01 quote(s(X)) -> s1(quote(activate(X))) quote(dbl(X)) -> dbl1(X) quote(sel(X, Y)) -> sel1(X, Y) s(X) -> n__s(X) dbl(X) -> n__dbl(X) dbls(X) -> n__dbls(X) sel(X1, X2) -> n__sel(X1, X2) indx(X1, X2) -> n__indx(X1, X2) from(X) -> n__from(X) activate(n__s(X)) -> s(X) activate(n__dbl(X)) -> dbl(activate(X)) activate(n__dbls(X)) -> dbls(activate(X)) activate(n__sel(X1, X2)) -> sel(activate(X1), activate(X2)) activate(n__indx(X1, X2)) -> indx(activate(X1), X2) activate(n__from(X)) -> from(X) activate(X) -> X 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 activate(n__indx(X1, X2)) ->^+ indx(activate(X1), X2) gives rise to a decreasing loop by considering the right hand sides subterm at position [0]. The pumping substitution is [X1 / n__indx(X1, X2)]. 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: dbl(0) -> 0 dbl(s(X)) -> s(n__s(n__dbl(activate(X)))) dbls(nil) -> nil dbls(cons(X, Y)) -> cons(n__dbl(activate(X)), n__dbls(activate(Y))) sel(0, cons(X, Y)) -> activate(X) sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z)) indx(nil, X) -> nil indx(cons(X, Y), Z) -> cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z))) from(X) -> cons(activate(X), n__from(n__s(activate(X)))) dbl1(0) -> 01 dbl1(s(X)) -> s1(s1(dbl1(activate(X)))) sel1(0, cons(X, Y)) -> activate(X) sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z)) quote(0) -> 01 quote(s(X)) -> s1(quote(activate(X))) quote(dbl(X)) -> dbl1(X) quote(sel(X, Y)) -> sel1(X, Y) s(X) -> n__s(X) dbl(X) -> n__dbl(X) dbls(X) -> n__dbls(X) sel(X1, X2) -> n__sel(X1, X2) indx(X1, X2) -> n__indx(X1, X2) from(X) -> n__from(X) activate(n__s(X)) -> s(X) activate(n__dbl(X)) -> dbl(activate(X)) activate(n__dbls(X)) -> dbls(activate(X)) activate(n__sel(X1, X2)) -> sel(activate(X1), activate(X2)) activate(n__indx(X1, X2)) -> indx(activate(X1), X2) activate(n__from(X)) -> from(X) activate(X) -> X 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: dbl(0) -> 0 dbl(s(X)) -> s(n__s(n__dbl(activate(X)))) dbls(nil) -> nil dbls(cons(X, Y)) -> cons(n__dbl(activate(X)), n__dbls(activate(Y))) sel(0, cons(X, Y)) -> activate(X) sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z)) indx(nil, X) -> nil indx(cons(X, Y), Z) -> cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z))) from(X) -> cons(activate(X), n__from(n__s(activate(X)))) dbl1(0) -> 01 dbl1(s(X)) -> s1(s1(dbl1(activate(X)))) sel1(0, cons(X, Y)) -> activate(X) sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z)) quote(0) -> 01 quote(s(X)) -> s1(quote(activate(X))) quote(dbl(X)) -> dbl1(X) quote(sel(X, Y)) -> sel1(X, Y) s(X) -> n__s(X) dbl(X) -> n__dbl(X) dbls(X) -> n__dbls(X) sel(X1, X2) -> n__sel(X1, X2) indx(X1, X2) -> n__indx(X1, X2) from(X) -> n__from(X) activate(n__s(X)) -> s(X) activate(n__dbl(X)) -> dbl(activate(X)) activate(n__dbls(X)) -> dbls(activate(X)) activate(n__sel(X1, X2)) -> sel(activate(X1), activate(X2)) activate(n__indx(X1, X2)) -> indx(activate(X1), X2) activate(n__from(X)) -> from(X) activate(X) -> X S is empty. Rewrite Strategy: FULL ---------------------------------------- (9) DecreasingLoopProof (FINISHED) The following loop(s) give(s) rise to the lower bound EXP: The rewrite sequence activate(n__from(X)) ->^+ cons(activate(X), n__from(n__s(activate(X)))) gives rise to a decreasing loop by considering the right hand sides subterm at position [0]. The pumping substitution is [X / n__from(X)]. The result substitution is [ ]. The rewrite sequence activate(n__from(X)) ->^+ cons(activate(X), n__from(n__s(activate(X)))) gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0,0]. The pumping substitution is [X / n__from(X)]. The result substitution is [ ]. ---------------------------------------- (10) BOUNDS(EXP, INF)