2.99/1.78 WORST_CASE(NON_POLY, ?) 2.99/1.79 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 2.99/1.79 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 2.99/1.79 2.99/1.79 2.99/1.79 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). 2.99/1.79 2.99/1.79 (0) CpxTRS 2.99/1.79 (1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] 2.99/1.79 (2) TRS for Loop Detection 2.99/1.79 (3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms] 2.99/1.79 (4) BEST 2.99/1.79 (5) proven lower bound 2.99/1.79 (6) LowerBoundPropagationProof [FINISHED, 0 ms] 2.99/1.79 (7) BOUNDS(n^1, INF) 2.99/1.79 (8) TRS for Loop Detection 2.99/1.79 (9) InfiniteLowerBoundProof [FINISHED, 0 ms] 2.99/1.79 (10) BOUNDS(INF, INF) 2.99/1.79 2.99/1.79 2.99/1.79 ---------------------------------------- 2.99/1.79 2.99/1.79 (0) 2.99/1.79 Obligation: 2.99/1.79 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). 2.99/1.79 2.99/1.79 2.99/1.79 The TRS R consists of the following rules: 2.99/1.79 2.99/1.79 filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M)) 2.99/1.79 filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M)) 2.99/1.79 sieve(cons(0, Y)) -> cons(0, sieve(Y)) 2.99/1.79 sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N))) 2.99/1.79 nats(N) -> cons(N, nats(s(N))) 2.99/1.79 zprimes -> sieve(nats(s(s(0)))) 2.99/1.79 2.99/1.79 S is empty. 2.99/1.79 Rewrite Strategy: INNERMOST 2.99/1.79 ---------------------------------------- 2.99/1.79 2.99/1.79 (1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID)) 2.99/1.79 Transformed a relative TRS into a decreasing-loop problem. 2.99/1.79 ---------------------------------------- 2.99/1.79 2.99/1.79 (2) 2.99/1.79 Obligation: 2.99/1.79 Analyzing the following TRS for decreasing loops: 2.99/1.79 2.99/1.79 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). 2.99/1.79 2.99/1.79 2.99/1.79 The TRS R consists of the following rules: 2.99/1.79 2.99/1.79 filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M)) 2.99/1.79 filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M)) 2.99/1.79 sieve(cons(0, Y)) -> cons(0, sieve(Y)) 2.99/1.79 sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N))) 2.99/1.79 nats(N) -> cons(N, nats(s(N))) 2.99/1.79 zprimes -> sieve(nats(s(s(0)))) 2.99/1.79 2.99/1.79 S is empty. 2.99/1.79 Rewrite Strategy: INNERMOST 2.99/1.79 ---------------------------------------- 2.99/1.79 2.99/1.79 (3) DecreasingLoopProof (LOWER BOUND(ID)) 2.99/1.79 The following loop(s) give(s) rise to the lower bound Omega(n^1): 2.99/1.79 2.99/1.79 The rewrite sequence 2.99/1.79 2.99/1.79 filter(cons(X, Y), s(N), M) ->^+ cons(X, filter(Y, N, M)) 2.99/1.79 2.99/1.79 gives rise to a decreasing loop by considering the right hand sides subterm at position [1]. 2.99/1.79 2.99/1.79 The pumping substitution is [Y / cons(X, Y), N / s(N)]. 2.99/1.79 2.99/1.79 The result substitution is [ ]. 2.99/1.79 2.99/1.79 2.99/1.79 2.99/1.79 2.99/1.79 ---------------------------------------- 2.99/1.79 2.99/1.79 (4) 2.99/1.79 Complex Obligation (BEST) 2.99/1.79 2.99/1.79 ---------------------------------------- 2.99/1.79 2.99/1.79 (5) 2.99/1.79 Obligation: 2.99/1.79 Proved the lower bound n^1 for the following obligation: 2.99/1.79 2.99/1.79 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). 2.99/1.79 2.99/1.79 2.99/1.79 The TRS R consists of the following rules: 2.99/1.79 2.99/1.79 filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M)) 2.99/1.79 filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M)) 2.99/1.79 sieve(cons(0, Y)) -> cons(0, sieve(Y)) 2.99/1.79 sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N))) 2.99/1.79 nats(N) -> cons(N, nats(s(N))) 2.99/1.79 zprimes -> sieve(nats(s(s(0)))) 2.99/1.79 2.99/1.79 S is empty. 2.99/1.79 Rewrite Strategy: INNERMOST 2.99/1.79 ---------------------------------------- 2.99/1.79 2.99/1.79 (6) LowerBoundPropagationProof (FINISHED) 2.99/1.79 Propagated lower bound. 2.99/1.79 ---------------------------------------- 2.99/1.79 2.99/1.79 (7) 2.99/1.79 BOUNDS(n^1, INF) 2.99/1.79 2.99/1.79 ---------------------------------------- 2.99/1.79 2.99/1.79 (8) 2.99/1.79 Obligation: 2.99/1.79 Analyzing the following TRS for decreasing loops: 2.99/1.79 2.99/1.79 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). 2.99/1.79 2.99/1.79 2.99/1.79 The TRS R consists of the following rules: 2.99/1.79 2.99/1.79 filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M)) 2.99/1.79 filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M)) 2.99/1.79 sieve(cons(0, Y)) -> cons(0, sieve(Y)) 2.99/1.79 sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N))) 2.99/1.79 nats(N) -> cons(N, nats(s(N))) 2.99/1.79 zprimes -> sieve(nats(s(s(0)))) 2.99/1.79 2.99/1.79 S is empty. 2.99/1.79 Rewrite Strategy: INNERMOST 2.99/1.79 ---------------------------------------- 2.99/1.79 2.99/1.79 (9) InfiniteLowerBoundProof (FINISHED) 2.99/1.79 The following loop proves infinite runtime complexity: 2.99/1.79 2.99/1.79 The rewrite sequence 2.99/1.79 2.99/1.79 nats(N) ->^+ cons(N, nats(s(N))) 2.99/1.79 2.99/1.79 gives rise to a decreasing loop by considering the right hand sides subterm at position [1]. 2.99/1.79 2.99/1.79 The pumping substitution is [ ]. 2.99/1.79 2.99/1.79 The result substitution is [N / s(N)]. 2.99/1.79 2.99/1.80 2.99/1.80 2.99/1.80 2.99/1.80 ---------------------------------------- 2.99/1.80 2.99/1.80 (10) 2.99/1.80 BOUNDS(INF, INF) 3.30/1.83 EOF