details

property	value
status	complete
benchmark	Ex6_15_AEL02_Z.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n147.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	2.92989 seconds
cpu usage	8.73013
user time	8.31741
system time	0.412726
max virtual memory	1.8279408E7
max residence set size	1234172.0

stage attributes

key	value
starexec-result	WORST_CASE(NON_POLY, ?)

output

WORST_CASE(NON_POLY, ?) 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).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) InfiniteLowerBoundProof [FINISHED, 858 ms] (10) BOUNDS(INF, INF) ---------------------------------------- (0) Obligation: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: sel(s(X), cons(Y, Z)) -> sel(X, activate(Z)) sel(0, cons(X, Z)) -> X first(0, Z) -> nil first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) from(X) -> cons(X, n__from(s(X))) sel1(s(X), cons(Y, Z)) -> sel1(X, activate(Z)) sel1(0, cons(X, Z)) -> quote(X) first1(0, Z) -> nil1 first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, activate(Z))) quote(n__0) -> 01 quote1(n__cons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z))) quote1(n__nil) -> nil1 quote(n__s(X)) -> s1(quote(activate(X))) quote(n__sel(X, Z)) -> sel1(activate(X), activate(Z)) quote1(n__first(X, Z)) -> first1(activate(X), activate(Z)) unquote(01) -> 0 unquote(s1(X)) -> s(unquote(X)) unquote1(nil1) -> nil unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z)) fcons(X, Z) -> cons(X, Z) first(X1, X2) -> n__first(X1, X2) from(X) -> n__from(X) 0 -> n__0 cons(X1, X2) -> n__cons(X1, X2) nil -> n__nil s(X) -> n__s(X) sel(X1, X2) -> n__sel(X1, X2) activate(n__first(X1, X2)) -> first(X1, X2) activate(n__from(X)) -> from(X) activate(n__0) -> 0 activate(n__cons(X1, X2)) -> cons(X1, X2) activate(n__nil) -> nil activate(n__s(X)) -> s(X) activate(n__sel(X1, X2)) -> sel(X1, X2) activate(X) -> 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(INF, INF). The TRS R consists of the following rules: sel(s(X), cons(Y, Z)) -> sel(X, activate(Z)) sel(0, cons(X, Z)) -> X first(0, Z) -> nil first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) from(X) -> cons(X, n__from(s(X))) sel1(s(X), cons(Y, Z)) -> sel1(X, activate(Z)) sel1(0, cons(X, Z)) -> quote(X) first1(0, Z) -> nil1 first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, activate(Z))) quote(n__0) -> 01 quote1(n__cons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z))) quote1(n__nil) -> nil1 quote(n__s(X)) -> s1(quote(activate(X))) quote(n__sel(X, Z)) -> sel1(activate(X), activate(Z)) quote1(n__first(X, Z)) -> first1(activate(X), activate(Z)) unquote(01) -> 0 unquote(s1(X)) -> s(unquote(X)) unquote1(nil1) -> nil popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to Runtime Complexity: TRS