/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(NON_POLY, ?) proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).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, 77 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: a__from(X) -> cons(mark(X), from(s(X))) a__2ndspos(0, Z) -> rnil a__2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(mark(Y)), a__2ndsneg(mark(N), mark(Z))) a__2ndsneg(0, Z) -> rnil a__2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(mark(Y)), a__2ndspos(mark(N), mark(Z))) a__pi(X) -> a__2ndspos(mark(X), a__from(0)) a__plus(0, Y) -> mark(Y) a__plus(s(X), Y) -> s(a__plus(mark(X), mark(Y))) a__times(0, Y) -> 0 a__times(s(X), Y) -> a__plus(mark(Y), a__times(mark(X), mark(Y))) a__square(X) -> a__times(mark(X), mark(X)) mark(from(X)) -> a__from(mark(X)) mark(2ndspos(X1, X2)) -> a__2ndspos(mark(X1), mark(X2)) mark(2ndsneg(X1, X2)) -> a__2ndsneg(mark(X1), mark(X2)) mark(pi(X)) -> a__pi(mark(X)) mark(plus(X1, X2)) -> a__plus(mark(X1), mark(X2)) mark(times(X1, X2)) -> a__times(mark(X1), mark(X2)) mark(square(X)) -> a__square(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(posrecip(X)) -> posrecip(mark(X)) mark(negrecip(X)) -> negrecip(mark(X)) mark(nil) -> nil mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(rnil) -> rnil mark(rcons(X1, X2)) -> rcons(mark(X1), mark(X2)) a__from(X) -> from(X) a__2ndspos(X1, X2) -> 2ndspos(X1, X2) a__2ndsneg(X1, X2) -> 2ndsneg(X1, X2) a__pi(X) -> pi(X) a__plus(X1, X2) -> plus(X1, X2) a__times(X1, X2) -> times(X1, X2) a__square(X) -> square(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: a__from(X) -> cons(mark(X), from(s(X))) a__2ndspos(0, Z) -> rnil a__2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(mark(Y)), a__2ndsneg(mark(N), mark(Z))) a__2ndsneg(0, Z) -> rnil a__2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(mark(Y)), a__2ndspos(mark(N), mark(Z))) a__pi(X) -> a__2ndspos(mark(X), a__from(0)) a__plus(0, Y) -> mark(Y) a__plus(s(X), Y) -> s(a__plus(mark(X), mark(Y))) a__times(0, Y) -> 0 a__times(s(X), Y) -> a__plus(mark(Y), a__times(mark(X), mark(Y))) a__square(X) -> a__times(mark(X), mark(X)) mark(from(X)) -> a__from(mark(X)) mark(2ndspos(X1, X2)) -> a__2ndspos(mark(X1), mark(X2)) mark(2ndsneg(X1, X2)) -> a__2ndsneg(mark(X1), mark(X2)) mark(pi(X)) -> a__pi(mark(X)) mark(plus(X1, X2)) -> a__plus(mark(X1), mark(X2)) mark(times(X1, X2)) -> a__times(mark(X1), mark(X2)) mark(square(X)) -> a__square(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(posrecip(X)) -> posrecip(mark(X)) mark(negrecip(X)) -> negrecip(mark(X)) mark(nil) -> nil mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(rnil) -> rnil mark(rcons(X1, X2)) -> rcons(mark(X1), mark(X2)) a__from(X) -> from(X) a__2ndspos(X1, X2) -> 2ndspos(X1, X2) a__2ndsneg(X1, X2) -> 2ndsneg(X1, X2) a__pi(X) -> pi(X) a__plus(X1, X2) -> plus(X1, X2) a__times(X1, X2) -> times(X1, X2) a__square(X) -> square(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(2ndsneg(X1, X2)) ->^+ a__2ndsneg(mark(X1), mark(X2)) gives rise to a decreasing loop by considering the right hand sides subterm at position [0]. The pumping substitution is [X1 / 2ndsneg(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: a__from(X) -> cons(mark(X), from(s(X))) a__2ndspos(0, Z) -> rnil a__2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(mark(Y)), a__2ndsneg(mark(N), mark(Z))) a__2ndsneg(0, Z) -> rnil a__2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(mark(Y)), a__2ndspos(mark(N), mark(Z))) a__pi(X) -> a__2ndspos(mark(X), a__from(0)) a__plus(0, Y) -> mark(Y) a__plus(s(X), Y) -> s(a__plus(mark(X), mark(Y))) a__times(0, Y) -> 0 a__times(s(X), Y) -> a__plus(mark(Y), a__times(mark(X), mark(Y))) a__square(X) -> a__times(mark(X), mark(X)) mark(from(X)) -> a__from(mark(X)) mark(2ndspos(X1, X2)) -> a__2ndspos(mark(X1), mark(X2)) mark(2ndsneg(X1, X2)) -> a__2ndsneg(mark(X1), mark(X2)) mark(pi(X)) -> a__pi(mark(X)) mark(plus(X1, X2)) -> a__plus(mark(X1), mark(X2)) mark(times(X1, X2)) -> a__times(mark(X1), mark(X2)) mark(square(X)) -> a__square(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(posrecip(X)) -> posrecip(mark(X)) mark(negrecip(X)) -> negrecip(mark(X)) mark(nil) -> nil mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(rnil) -> rnil mark(rcons(X1, X2)) -> rcons(mark(X1), mark(X2)) a__from(X) -> from(X) a__2ndspos(X1, X2) -> 2ndspos(X1, X2) a__2ndsneg(X1, X2) -> 2ndsneg(X1, X2) a__pi(X) -> pi(X) a__plus(X1, X2) -> plus(X1, X2) a__times(X1, X2) -> times(X1, X2) a__square(X) -> square(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: a__from(X) -> cons(mark(X), from(s(X))) a__2ndspos(0, Z) -> rnil a__2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(mark(Y)), a__2ndsneg(mark(N), mark(Z))) a__2ndsneg(0, Z) -> rnil a__2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(mark(Y)), a__2ndspos(mark(N), mark(Z))) a__pi(X) -> a__2ndspos(mark(X), a__from(0)) a__plus(0, Y) -> mark(Y) a__plus(s(X), Y) -> s(a__plus(mark(X), mark(Y))) a__times(0, Y) -> 0 a__times(s(X), Y) -> a__plus(mark(Y), a__times(mark(X), mark(Y))) a__square(X) -> a__times(mark(X), mark(X)) mark(from(X)) -> a__from(mark(X)) mark(2ndspos(X1, X2)) -> a__2ndspos(mark(X1), mark(X2)) mark(2ndsneg(X1, X2)) -> a__2ndsneg(mark(X1), mark(X2)) mark(pi(X)) -> a__pi(mark(X)) mark(plus(X1, X2)) -> a__plus(mark(X1), mark(X2)) mark(times(X1, X2)) -> a__times(mark(X1), mark(X2)) mark(square(X)) -> a__square(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(posrecip(X)) -> posrecip(mark(X)) mark(negrecip(X)) -> negrecip(mark(X)) mark(nil) -> nil mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(rnil) -> rnil mark(rcons(X1, X2)) -> rcons(mark(X1), mark(X2)) a__from(X) -> from(X) a__2ndspos(X1, X2) -> 2ndspos(X1, X2) a__2ndsneg(X1, X2) -> 2ndsneg(X1, X2) a__pi(X) -> pi(X) a__plus(X1, X2) -> plus(X1, X2) a__times(X1, X2) -> times(X1, X2) a__square(X) -> square(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 mark(from(X)) ->^+ cons(mark(mark(X)), from(s(mark(X)))) gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0]. The pumping substitution is [X / from(X)]. The result substitution is [ ]. The rewrite sequence mark(from(X)) ->^+ cons(mark(mark(X)), from(s(mark(X)))) gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0,0]. The pumping substitution is [X / from(X)]. The result substitution is [ ]. ---------------------------------------- (10) BOUNDS(EXP, INF)