details

property	value
status	complete
benchmark	ex2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n139.star.cs.uiowa.edu
space	Strategy_removed_mixed_05

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	1.50869 seconds
cpu usage	3.13006
user time	2.99112
system time	0.138935
max virtual memory	1.8275312E7
max residence set size	211652.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).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) InfiniteLowerBoundProof [FINISHED, 0 ms] (4) 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: cons(x, cons(y, z)) -> big inf(x) -> cons(x, inf(s(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: cons(x, cons(y, z)) -> big inf(x) -> cons(x, inf(s(x))) S is empty. Rewrite Strategy: FULL ---------------------------------------- (3) InfiniteLowerBoundProof (FINISHED) The following loop proves infinite runtime complexity: The rewrite sequence inf(x) ->^+ cons(x, inf(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

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job log

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all output return to Runtime Complexity: TRS