details

property	value
status	complete
benchmark	Ex6_9_Luc02c.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n138.star.cs.uiowa.edu
space	Strategy_removed_CSR_05

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	4.46061 seconds
cpu usage	3.4348
user time	3.29223
system time	0.142575
max virtual memory	1.8273228E7
max residence set size	215836.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF).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) InfiniteLowerBoundProof [FINISHED, 0 ms] (4) BOUNDS(INF, INF) ---------------------------------------- (0) Obligation: The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: 2nd(cons1(X, cons(Y, Z))) -> Y 2nd(cons(X, X1)) -> 2nd(cons1(X, X1)) from(X) -> cons(X, from(s(X))) S is empty. Rewrite Strategy: INNERMOST ---------------------------------------- (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 (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: 2nd(cons1(X, cons(Y, Z))) -> Y 2nd(cons(X, X1)) -> 2nd(cons1(X, X1)) from(X) -> cons(X, from(s(X))) S is empty. Rewrite Strategy: INNERMOST ---------------------------------------- (3) InfiniteLowerBoundProof (FINISHED) The following loop proves infinite runtime complexity: The rewrite sequence from(X) ->^+ cons(X, from(s(X))) gives rise to a decreasing loop by considering the right hand sides subterm at position [1]. The pumping substitution is [ ]. The result substitution is [X / s(X)]. ---------------------------------------- (4) BOUNDS(INF, INF) popout

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

job log

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actions

all output return to Runtime Complexity: TRS Innermost