3.38/1.67 WORST_CASE(NON_POLY, ?) 3.38/1.68 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 3.38/1.68 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.38/1.68 3.38/1.68 3.38/1.68 The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). 3.38/1.68 3.38/1.68 (0) CpxTRS 3.38/1.68 (1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] 3.38/1.68 (2) TRS for Loop Detection 3.38/1.68 (3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms] 3.38/1.68 (4) BEST 3.38/1.68 (5) proven lower bound 3.38/1.68 (6) LowerBoundPropagationProof [FINISHED, 0 ms] 3.38/1.68 (7) BOUNDS(n^1, INF) 3.38/1.68 (8) TRS for Loop Detection 3.38/1.68 (9) DecreasingLoopProof [FINISHED, 22 ms] 3.38/1.68 (10) BOUNDS(EXP, INF) 3.38/1.68 3.38/1.68 3.38/1.68 ---------------------------------------- 3.38/1.68 3.38/1.68 (0) 3.38/1.68 Obligation: 3.38/1.68 The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). 3.38/1.68 3.38/1.68 3.38/1.68 The TRS R consists of the following rules: 3.38/1.68 3.38/1.68 primes -> sieve(from(s(s(0)))) 3.38/1.68 from(X) -> cons(X, n__from(n__s(X))) 3.38/1.68 head(cons(X, Y)) -> X 3.38/1.68 tail(cons(X, Y)) -> activate(Y) 3.38/1.68 if(true, X, Y) -> activate(X) 3.38/1.68 if(false, X, Y) -> activate(Y) 3.38/1.68 filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), n__filter(n__s(n__s(X)), activate(Z)), n__cons(Y, n__filter(X, n__sieve(Y)))) 3.38/1.68 sieve(cons(X, Y)) -> cons(X, n__filter(X, n__sieve(activate(Y)))) 3.38/1.68 from(X) -> n__from(X) 3.38/1.68 s(X) -> n__s(X) 3.38/1.68 filter(X1, X2) -> n__filter(X1, X2) 3.38/1.68 cons(X1, X2) -> n__cons(X1, X2) 3.38/1.68 sieve(X) -> n__sieve(X) 3.38/1.68 activate(n__from(X)) -> from(activate(X)) 3.38/1.68 activate(n__s(X)) -> s(activate(X)) 3.38/1.68 activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2)) 3.38/1.68 activate(n__cons(X1, X2)) -> cons(activate(X1), X2) 3.38/1.68 activate(n__sieve(X)) -> sieve(activate(X)) 3.38/1.68 activate(X) -> X 3.38/1.68 3.38/1.68 S is empty. 3.38/1.68 Rewrite Strategy: FULL 3.38/1.68 ---------------------------------------- 3.38/1.68 3.38/1.68 (1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID)) 3.38/1.68 Transformed a relative TRS into a decreasing-loop problem. 3.38/1.68 ---------------------------------------- 3.38/1.68 3.38/1.68 (2) 3.38/1.68 Obligation: 3.38/1.68 Analyzing the following TRS for decreasing loops: 3.38/1.68 3.38/1.68 The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). 3.38/1.68 3.38/1.68 3.38/1.68 The TRS R consists of the following rules: 3.38/1.68 3.38/1.68 primes -> sieve(from(s(s(0)))) 3.38/1.68 from(X) -> cons(X, n__from(n__s(X))) 3.38/1.68 head(cons(X, Y)) -> X 3.38/1.68 tail(cons(X, Y)) -> activate(Y) 3.38/1.68 if(true, X, Y) -> activate(X) 3.38/1.68 if(false, X, Y) -> activate(Y) 3.38/1.68 filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), n__filter(n__s(n__s(X)), activate(Z)), n__cons(Y, n__filter(X, n__sieve(Y)))) 3.38/1.68 sieve(cons(X, Y)) -> cons(X, n__filter(X, n__sieve(activate(Y)))) 3.38/1.68 from(X) -> n__from(X) 3.38/1.68 s(X) -> n__s(X) 3.38/1.68 filter(X1, X2) -> n__filter(X1, X2) 3.38/1.68 cons(X1, X2) -> n__cons(X1, X2) 3.38/1.68 sieve(X) -> n__sieve(X) 3.38/1.68 activate(n__from(X)) -> from(activate(X)) 3.38/1.68 activate(n__s(X)) -> s(activate(X)) 3.38/1.68 activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2)) 3.38/1.68 activate(n__cons(X1, X2)) -> cons(activate(X1), X2) 3.38/1.68 activate(n__sieve(X)) -> sieve(activate(X)) 3.38/1.68 activate(X) -> X 3.38/1.68 3.38/1.68 S is empty. 3.38/1.68 Rewrite Strategy: FULL 3.38/1.68 ---------------------------------------- 3.38/1.68 3.38/1.68 (3) DecreasingLoopProof (LOWER BOUND(ID)) 3.38/1.68 The following loop(s) give(s) rise to the lower bound Omega(n^1): 3.38/1.68 3.38/1.68 The rewrite sequence 3.38/1.68 3.38/1.68 activate(n__s(X)) ->^+ s(activate(X)) 3.38/1.68 3.38/1.68 gives rise to a decreasing loop by considering the right hand sides subterm at position [0]. 3.38/1.68 3.38/1.68 The pumping substitution is [X / n__s(X)]. 3.38/1.68 3.38/1.68 The result substitution is [ ]. 3.38/1.68 3.38/1.68 3.38/1.68 3.38/1.68 3.38/1.68 ---------------------------------------- 3.38/1.68 3.38/1.68 (4) 3.38/1.68 Complex Obligation (BEST) 3.38/1.68 3.38/1.68 ---------------------------------------- 3.38/1.68 3.38/1.68 (5) 3.38/1.68 Obligation: 3.38/1.68 Proved the lower bound n^1 for the following obligation: 3.38/1.68 3.38/1.68 The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). 3.38/1.68 3.38/1.68 3.38/1.68 The TRS R consists of the following rules: 3.38/1.68 3.38/1.68 primes -> sieve(from(s(s(0)))) 3.38/1.68 from(X) -> cons(X, n__from(n__s(X))) 3.38/1.68 head(cons(X, Y)) -> X 3.38/1.68 tail(cons(X, Y)) -> activate(Y) 3.38/1.68 if(true, X, Y) -> activate(X) 3.38/1.68 if(false, X, Y) -> activate(Y) 3.38/1.68 filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), n__filter(n__s(n__s(X)), activate(Z)), n__cons(Y, n__filter(X, n__sieve(Y)))) 3.38/1.68 sieve(cons(X, Y)) -> cons(X, n__filter(X, n__sieve(activate(Y)))) 3.38/1.68 from(X) -> n__from(X) 3.38/1.68 s(X) -> n__s(X) 3.38/1.68 filter(X1, X2) -> n__filter(X1, X2) 3.38/1.68 cons(X1, X2) -> n__cons(X1, X2) 3.38/1.68 sieve(X) -> n__sieve(X) 3.38/1.68 activate(n__from(X)) -> from(activate(X)) 3.38/1.68 activate(n__s(X)) -> s(activate(X)) 3.38/1.68 activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2)) 3.38/1.68 activate(n__cons(X1, X2)) -> cons(activate(X1), X2) 3.38/1.68 activate(n__sieve(X)) -> sieve(activate(X)) 3.38/1.68 activate(X) -> X 3.38/1.68 3.38/1.68 S is empty. 3.38/1.68 Rewrite Strategy: FULL 3.38/1.68 ---------------------------------------- 3.38/1.68 3.38/1.68 (6) LowerBoundPropagationProof (FINISHED) 3.38/1.68 Propagated lower bound. 3.38/1.68 ---------------------------------------- 3.38/1.68 3.38/1.68 (7) 3.38/1.68 BOUNDS(n^1, INF) 3.38/1.68 3.38/1.68 ---------------------------------------- 3.38/1.68 3.38/1.68 (8) 3.38/1.68 Obligation: 3.38/1.68 Analyzing the following TRS for decreasing loops: 3.38/1.68 3.38/1.68 The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). 3.38/1.68 3.38/1.68 3.38/1.68 The TRS R consists of the following rules: 3.38/1.68 3.38/1.68 primes -> sieve(from(s(s(0)))) 3.38/1.68 from(X) -> cons(X, n__from(n__s(X))) 3.38/1.68 head(cons(X, Y)) -> X 3.38/1.68 tail(cons(X, Y)) -> activate(Y) 3.38/1.68 if(true, X, Y) -> activate(X) 3.38/1.68 if(false, X, Y) -> activate(Y) 3.38/1.68 filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), n__filter(n__s(n__s(X)), activate(Z)), n__cons(Y, n__filter(X, n__sieve(Y)))) 3.38/1.68 sieve(cons(X, Y)) -> cons(X, n__filter(X, n__sieve(activate(Y)))) 3.38/1.68 from(X) -> n__from(X) 3.38/1.68 s(X) -> n__s(X) 3.38/1.68 filter(X1, X2) -> n__filter(X1, X2) 3.38/1.68 cons(X1, X2) -> n__cons(X1, X2) 3.38/1.68 sieve(X) -> n__sieve(X) 3.38/1.68 activate(n__from(X)) -> from(activate(X)) 3.38/1.68 activate(n__s(X)) -> s(activate(X)) 3.38/1.68 activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2)) 3.38/1.68 activate(n__cons(X1, X2)) -> cons(activate(X1), X2) 3.38/1.68 activate(n__sieve(X)) -> sieve(activate(X)) 3.38/1.68 activate(X) -> X 3.38/1.68 3.38/1.68 S is empty. 3.38/1.68 Rewrite Strategy: FULL 3.38/1.68 ---------------------------------------- 3.38/1.68 3.38/1.68 (9) DecreasingLoopProof (FINISHED) 3.38/1.68 The following loop(s) give(s) rise to the lower bound EXP: 3.38/1.68 3.38/1.68 The rewrite sequence 3.38/1.68 3.38/1.68 activate(n__from(X)) ->^+ cons(activate(X), n__from(n__s(activate(X)))) 3.38/1.68 3.38/1.68 gives rise to a decreasing loop by considering the right hand sides subterm at position [0]. 3.38/1.68 3.38/1.68 The pumping substitution is [X / n__from(X)]. 3.38/1.68 3.38/1.68 The result substitution is [ ]. 3.38/1.68 3.38/1.68 3.38/1.68 3.38/1.68 The rewrite sequence 3.38/1.68 3.38/1.68 activate(n__from(X)) ->^+ cons(activate(X), n__from(n__s(activate(X)))) 3.38/1.68 3.38/1.68 gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0,0]. 3.38/1.68 3.38/1.68 The pumping substitution is [X / n__from(X)]. 3.38/1.68 3.38/1.68 The result substitution is [ ]. 3.38/1.68 3.38/1.68 3.38/1.68 3.38/1.68 3.38/1.68 ---------------------------------------- 3.38/1.68 3.38/1.68 (10) 3.38/1.68 BOUNDS(EXP, INF) 3.66/1.72 EOF