3.40/1.71 WORST_CASE(NON_POLY, ?) 3.40/1.72 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 3.40/1.72 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.40/1.72 3.40/1.72 3.40/1.72 The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). 3.40/1.72 3.40/1.72 (0) CpxTRS 3.40/1.72 (1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] 3.40/1.72 (2) TRS for Loop Detection 3.40/1.72 (3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms] 3.40/1.72 (4) BEST 3.40/1.72 (5) proven lower bound 3.40/1.72 (6) LowerBoundPropagationProof [FINISHED, 0 ms] 3.40/1.72 (7) BOUNDS(n^1, INF) 3.40/1.72 (8) TRS for Loop Detection 3.40/1.72 (9) DecreasingLoopProof [FINISHED, 23 ms] 3.40/1.72 (10) BOUNDS(EXP, INF) 3.40/1.72 3.40/1.72 3.40/1.72 ---------------------------------------- 3.40/1.72 3.40/1.72 (0) 3.40/1.72 Obligation: 3.40/1.72 The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). 3.40/1.72 3.40/1.72 3.40/1.72 The TRS R consists of the following rules: 3.40/1.72 3.40/1.72 a__filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M)) 3.40/1.72 a__filter(cons(X, Y), s(N), M) -> cons(mark(X), filter(Y, N, M)) 3.40/1.72 a__sieve(cons(0, Y)) -> cons(0, sieve(Y)) 3.40/1.72 a__sieve(cons(s(N), Y)) -> cons(s(mark(N)), sieve(filter(Y, N, N))) 3.40/1.72 a__nats(N) -> cons(mark(N), nats(s(N))) 3.40/1.72 a__zprimes -> a__sieve(a__nats(s(s(0)))) 3.40/1.72 mark(filter(X1, X2, X3)) -> a__filter(mark(X1), mark(X2), mark(X3)) 3.40/1.72 mark(sieve(X)) -> a__sieve(mark(X)) 3.40/1.72 mark(nats(X)) -> a__nats(mark(X)) 3.40/1.72 mark(zprimes) -> a__zprimes 3.40/1.72 mark(cons(X1, X2)) -> cons(mark(X1), X2) 3.40/1.72 mark(0) -> 0 3.40/1.72 mark(s(X)) -> s(mark(X)) 3.40/1.72 a__filter(X1, X2, X3) -> filter(X1, X2, X3) 3.40/1.72 a__sieve(X) -> sieve(X) 3.40/1.72 a__nats(X) -> nats(X) 3.40/1.72 a__zprimes -> zprimes 3.40/1.72 3.40/1.72 S is empty. 3.40/1.72 Rewrite Strategy: FULL 3.40/1.72 ---------------------------------------- 3.40/1.72 3.40/1.72 (1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID)) 3.40/1.72 Transformed a relative TRS into a decreasing-loop problem. 3.40/1.72 ---------------------------------------- 3.40/1.72 3.40/1.72 (2) 3.40/1.72 Obligation: 3.40/1.72 Analyzing the following TRS for decreasing loops: 3.40/1.72 3.40/1.72 The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). 3.40/1.72 3.40/1.72 3.40/1.72 The TRS R consists of the following rules: 3.40/1.72 3.40/1.72 a__filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M)) 3.40/1.72 a__filter(cons(X, Y), s(N), M) -> cons(mark(X), filter(Y, N, M)) 3.40/1.72 a__sieve(cons(0, Y)) -> cons(0, sieve(Y)) 3.40/1.72 a__sieve(cons(s(N), Y)) -> cons(s(mark(N)), sieve(filter(Y, N, N))) 3.40/1.72 a__nats(N) -> cons(mark(N), nats(s(N))) 3.40/1.72 a__zprimes -> a__sieve(a__nats(s(s(0)))) 3.40/1.72 mark(filter(X1, X2, X3)) -> a__filter(mark(X1), mark(X2), mark(X3)) 3.40/1.72 mark(sieve(X)) -> a__sieve(mark(X)) 3.40/1.72 mark(nats(X)) -> a__nats(mark(X)) 3.40/1.72 mark(zprimes) -> a__zprimes 3.40/1.72 mark(cons(X1, X2)) -> cons(mark(X1), X2) 3.40/1.72 mark(0) -> 0 3.40/1.72 mark(s(X)) -> s(mark(X)) 3.40/1.72 a__filter(X1, X2, X3) -> filter(X1, X2, X3) 3.40/1.72 a__sieve(X) -> sieve(X) 3.40/1.72 a__nats(X) -> nats(X) 3.40/1.72 a__zprimes -> zprimes 3.40/1.72 3.40/1.72 S is empty. 3.40/1.72 Rewrite Strategy: FULL 3.40/1.72 ---------------------------------------- 3.40/1.72 3.40/1.72 (3) DecreasingLoopProof (LOWER BOUND(ID)) 3.40/1.72 The following loop(s) give(s) rise to the lower bound Omega(n^1): 3.40/1.72 3.40/1.72 The rewrite sequence 3.40/1.72 3.40/1.72 mark(nats(X)) ->^+ a__nats(mark(X)) 3.40/1.72 3.40/1.72 gives rise to a decreasing loop by considering the right hand sides subterm at position [0]. 3.40/1.72 3.40/1.72 The pumping substitution is [X / nats(X)]. 3.40/1.72 3.40/1.72 The result substitution is [ ]. 3.40/1.72 3.40/1.72 3.40/1.72 3.40/1.72 3.40/1.72 ---------------------------------------- 3.40/1.72 3.40/1.72 (4) 3.40/1.72 Complex Obligation (BEST) 3.40/1.72 3.40/1.72 ---------------------------------------- 3.40/1.72 3.40/1.72 (5) 3.40/1.72 Obligation: 3.40/1.72 Proved the lower bound n^1 for the following obligation: 3.40/1.72 3.40/1.72 The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). 3.40/1.72 3.40/1.72 3.40/1.72 The TRS R consists of the following rules: 3.40/1.72 3.40/1.72 a__filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M)) 3.40/1.72 a__filter(cons(X, Y), s(N), M) -> cons(mark(X), filter(Y, N, M)) 3.40/1.72 a__sieve(cons(0, Y)) -> cons(0, sieve(Y)) 3.40/1.72 a__sieve(cons(s(N), Y)) -> cons(s(mark(N)), sieve(filter(Y, N, N))) 3.40/1.72 a__nats(N) -> cons(mark(N), nats(s(N))) 3.40/1.72 a__zprimes -> a__sieve(a__nats(s(s(0)))) 3.40/1.72 mark(filter(X1, X2, X3)) -> a__filter(mark(X1), mark(X2), mark(X3)) 3.40/1.72 mark(sieve(X)) -> a__sieve(mark(X)) 3.40/1.72 mark(nats(X)) -> a__nats(mark(X)) 3.40/1.72 mark(zprimes) -> a__zprimes 3.40/1.72 mark(cons(X1, X2)) -> cons(mark(X1), X2) 3.40/1.72 mark(0) -> 0 3.40/1.72 mark(s(X)) -> s(mark(X)) 3.40/1.72 a__filter(X1, X2, X3) -> filter(X1, X2, X3) 3.40/1.72 a__sieve(X) -> sieve(X) 3.40/1.72 a__nats(X) -> nats(X) 3.40/1.72 a__zprimes -> zprimes 3.40/1.72 3.40/1.72 S is empty. 3.40/1.72 Rewrite Strategy: FULL 3.40/1.72 ---------------------------------------- 3.40/1.72 3.40/1.72 (6) LowerBoundPropagationProof (FINISHED) 3.40/1.72 Propagated lower bound. 3.40/1.72 ---------------------------------------- 3.40/1.72 3.40/1.72 (7) 3.40/1.72 BOUNDS(n^1, INF) 3.40/1.72 3.40/1.72 ---------------------------------------- 3.40/1.72 3.40/1.72 (8) 3.40/1.72 Obligation: 3.40/1.72 Analyzing the following TRS for decreasing loops: 3.40/1.72 3.40/1.72 The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). 3.40/1.72 3.40/1.72 3.40/1.72 The TRS R consists of the following rules: 3.40/1.72 3.40/1.72 a__filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M)) 3.40/1.72 a__filter(cons(X, Y), s(N), M) -> cons(mark(X), filter(Y, N, M)) 3.40/1.72 a__sieve(cons(0, Y)) -> cons(0, sieve(Y)) 3.40/1.72 a__sieve(cons(s(N), Y)) -> cons(s(mark(N)), sieve(filter(Y, N, N))) 3.40/1.72 a__nats(N) -> cons(mark(N), nats(s(N))) 3.40/1.72 a__zprimes -> a__sieve(a__nats(s(s(0)))) 3.40/1.72 mark(filter(X1, X2, X3)) -> a__filter(mark(X1), mark(X2), mark(X3)) 3.40/1.72 mark(sieve(X)) -> a__sieve(mark(X)) 3.40/1.72 mark(nats(X)) -> a__nats(mark(X)) 3.40/1.72 mark(zprimes) -> a__zprimes 3.40/1.72 mark(cons(X1, X2)) -> cons(mark(X1), X2) 3.40/1.72 mark(0) -> 0 3.40/1.72 mark(s(X)) -> s(mark(X)) 3.40/1.72 a__filter(X1, X2, X3) -> filter(X1, X2, X3) 3.40/1.72 a__sieve(X) -> sieve(X) 3.40/1.72 a__nats(X) -> nats(X) 3.40/1.72 a__zprimes -> zprimes 3.40/1.72 3.40/1.72 S is empty. 3.40/1.72 Rewrite Strategy: FULL 3.40/1.72 ---------------------------------------- 3.40/1.72 3.40/1.72 (9) DecreasingLoopProof (FINISHED) 3.40/1.72 The following loop(s) give(s) rise to the lower bound EXP: 3.40/1.72 3.40/1.72 The rewrite sequence 3.40/1.72 3.40/1.72 mark(nats(X)) ->^+ cons(mark(mark(X)), nats(s(mark(X)))) 3.40/1.72 3.40/1.72 gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0]. 3.40/1.72 3.40/1.72 The pumping substitution is [X / nats(X)]. 3.40/1.72 3.40/1.72 The result substitution is [ ]. 3.40/1.72 3.40/1.72 3.40/1.72 3.40/1.72 The rewrite sequence 3.40/1.72 3.40/1.72 mark(nats(X)) ->^+ cons(mark(mark(X)), nats(s(mark(X)))) 3.40/1.72 3.40/1.72 gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0,0]. 3.40/1.72 3.40/1.72 The pumping substitution is [X / nats(X)]. 3.40/1.72 3.40/1.72 The result substitution is [ ]. 3.40/1.72 3.40/1.72 3.40/1.72 3.40/1.72 3.40/1.72 ---------------------------------------- 3.40/1.72 3.40/1.72 (10) 3.40/1.72 BOUNDS(EXP, INF) 3.73/1.75 EOF