/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^1), ?) 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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 ---------------------------------------- (0) Obligation: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF). The TRS R consists of the following rules: a__dbl(0) -> 0 a__dbl(s(X)) -> s(s(dbl(X))) a__dbls(nil) -> nil a__dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y)) a__sel(0, cons(X, Y)) -> mark(X) a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z)) a__indx(nil, X) -> nil a__indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z)) a__from(X) -> cons(X, from(s(X))) a__dbl1(0) -> 01 a__dbl1(s(X)) -> s1(s1(a__dbl1(mark(X)))) a__sel1(0, cons(X, Y)) -> mark(X) a__sel1(s(X), cons(Y, Z)) -> a__sel1(mark(X), mark(Z)) a__quote(0) -> 01 a__quote(s(X)) -> s1(a__quote(mark(X))) a__quote(dbl(X)) -> a__dbl1(mark(X)) a__quote(sel(X, Y)) -> a__sel1(mark(X), mark(Y)) mark(dbl(X)) -> a__dbl(mark(X)) mark(dbls(X)) -> a__dbls(mark(X)) mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2)) mark(indx(X1, X2)) -> a__indx(mark(X1), X2) mark(from(X)) -> a__from(X) mark(dbl1(X)) -> a__dbl1(mark(X)) mark(sel1(X1, X2)) -> a__sel1(mark(X1), mark(X2)) mark(quote(X)) -> a__quote(mark(X)) mark(0) -> 0 mark(s(X)) -> s(X) mark(nil) -> nil mark(cons(X1, X2)) -> cons(X1, X2) mark(01) -> 01 mark(s1(X)) -> s1(mark(X)) a__dbl(X) -> dbl(X) a__dbls(X) -> dbls(X) a__sel(X1, X2) -> sel(X1, X2) a__indx(X1, X2) -> indx(X1, X2) a__from(X) -> from(X) a__dbl1(X) -> dbl1(X) a__sel1(X1, X2) -> sel1(X1, X2) a__quote(X) -> quote(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(n^1, INF). The TRS R consists of the following rules: a__dbl(0) -> 0 a__dbl(s(X)) -> s(s(dbl(X))) a__dbls(nil) -> nil a__dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y)) a__sel(0, cons(X, Y)) -> mark(X) a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z)) a__indx(nil, X) -> nil a__indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z)) a__from(X) -> cons(X, from(s(X))) a__dbl1(0) -> 01 a__dbl1(s(X)) -> s1(s1(a__dbl1(mark(X)))) a__sel1(0, cons(X, Y)) -> mark(X) a__sel1(s(X), cons(Y, Z)) -> a__sel1(mark(X), mark(Z)) a__quote(0) -> 01 a__quote(s(X)) -> s1(a__quote(mark(X))) a__quote(dbl(X)) -> a__dbl1(mark(X)) a__quote(sel(X, Y)) -> a__sel1(mark(X), mark(Y)) mark(dbl(X)) -> a__dbl(mark(X)) mark(dbls(X)) -> a__dbls(mark(X)) mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2)) mark(indx(X1, X2)) -> a__indx(mark(X1), X2) mark(from(X)) -> a__from(X) mark(dbl1(X)) -> a__dbl1(mark(X)) mark(sel1(X1, X2)) -> a__sel1(mark(X1), mark(X2)) mark(quote(X)) -> a__quote(mark(X)) mark(0) -> 0 mark(s(X)) -> s(X) mark(nil) -> nil mark(cons(X1, X2)) -> cons(X1, X2) mark(01) -> 01 mark(s1(X)) -> s1(mark(X)) a__dbl(X) -> dbl(X) a__dbls(X) -> dbls(X) a__sel(X1, X2) -> sel(X1, X2) a__indx(X1, X2) -> indx(X1, X2) a__from(X) -> from(X) a__dbl1(X) -> dbl1(X) a__sel1(X1, X2) -> sel1(X1, X2) a__quote(X) -> quote(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 mark(indx(X1, X2)) ->^+ a__indx(mark(X1), X2) gives rise to a decreasing loop by considering the right hand sides subterm at position [0]. The pumping substitution is [X1 / 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(n^1, INF). The TRS R consists of the following rules: a__dbl(0) -> 0 a__dbl(s(X)) -> s(s(dbl(X))) a__dbls(nil) -> nil a__dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y)) a__sel(0, cons(X, Y)) -> mark(X) a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z)) a__indx(nil, X) -> nil a__indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z)) a__from(X) -> cons(X, from(s(X))) a__dbl1(0) -> 01 a__dbl1(s(X)) -> s1(s1(a__dbl1(mark(X)))) a__sel1(0, cons(X, Y)) -> mark(X) a__sel1(s(X), cons(Y, Z)) -> a__sel1(mark(X), mark(Z)) a__quote(0) -> 01 a__quote(s(X)) -> s1(a__quote(mark(X))) a__quote(dbl(X)) -> a__dbl1(mark(X)) a__quote(sel(X, Y)) -> a__sel1(mark(X), mark(Y)) mark(dbl(X)) -> a__dbl(mark(X)) mark(dbls(X)) -> a__dbls(mark(X)) mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2)) mark(indx(X1, X2)) -> a__indx(mark(X1), X2) mark(from(X)) -> a__from(X) mark(dbl1(X)) -> a__dbl1(mark(X)) mark(sel1(X1, X2)) -> a__sel1(mark(X1), mark(X2)) mark(quote(X)) -> a__quote(mark(X)) mark(0) -> 0 mark(s(X)) -> s(X) mark(nil) -> nil mark(cons(X1, X2)) -> cons(X1, X2) mark(01) -> 01 mark(s1(X)) -> s1(mark(X)) a__dbl(X) -> dbl(X) a__dbls(X) -> dbls(X) a__sel(X1, X2) -> sel(X1, X2) a__indx(X1, X2) -> indx(X1, X2) a__from(X) -> from(X) a__dbl1(X) -> dbl1(X) a__sel1(X1, X2) -> sel1(X1, X2) a__quote(X) -> quote(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(n^1, INF). The TRS R consists of the following rules: a__dbl(0) -> 0 a__dbl(s(X)) -> s(s(dbl(X))) a__dbls(nil) -> nil a__dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y)) a__sel(0, cons(X, Y)) -> mark(X) a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z)) a__indx(nil, X) -> nil a__indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z)) a__from(X) -> cons(X, from(s(X))) a__dbl1(0) -> 01 a__dbl1(s(X)) -> s1(s1(a__dbl1(mark(X)))) a__sel1(0, cons(X, Y)) -> mark(X) a__sel1(s(X), cons(Y, Z)) -> a__sel1(mark(X), mark(Z)) a__quote(0) -> 01 a__quote(s(X)) -> s1(a__quote(mark(X))) a__quote(dbl(X)) -> a__dbl1(mark(X)) a__quote(sel(X, Y)) -> a__sel1(mark(X), mark(Y)) mark(dbl(X)) -> a__dbl(mark(X)) mark(dbls(X)) -> a__dbls(mark(X)) mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2)) mark(indx(X1, X2)) -> a__indx(mark(X1), X2) mark(from(X)) -> a__from(X) mark(dbl1(X)) -> a__dbl1(mark(X)) mark(sel1(X1, X2)) -> a__sel1(mark(X1), mark(X2)) mark(quote(X)) -> a__quote(mark(X)) mark(0) -> 0 mark(s(X)) -> s(X) mark(nil) -> nil mark(cons(X1, X2)) -> cons(X1, X2) mark(01) -> 01 mark(s1(X)) -> s1(mark(X)) a__dbl(X) -> dbl(X) a__dbls(X) -> dbls(X) a__sel(X1, X2) -> sel(X1, X2) a__indx(X1, X2) -> indx(X1, X2) a__from(X) -> from(X) a__dbl1(X) -> dbl1(X) a__sel1(X1, X2) -> sel1(X1, X2) a__quote(X) -> quote(X) S is empty. Rewrite Strategy: FULL