details

property	value
status	complete
benchmark	letexp.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	Frederiksen_Glenstrup

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	1.52553 seconds
cpu usage	3.454
user time	3.31873
system time	0.135273
max virtual memory	1.8275312E7
max residence set size	224724.0

stage attributes

key	value
starexec-result	WORST_CASE(NON_POLY, ?)

output

WORST_CASE(NON_POLY, ?) proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF).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: letexp(x, y) -> letexp(Cons(Cons(Nil, Nil), x), y) goal(x, y) -> letexp(x, y) 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: letexp(x, y) -> letexp(Cons(Cons(Nil, Nil), x), y) goal(x, y) -> letexp(x, y) S is empty. Rewrite Strategy: INNERMOST ---------------------------------------- (3) InfiniteLowerBoundProof (FINISHED) The following loop proves infinite runtime complexity: The rewrite sequence letexp(x, y) ->^+ letexp(Cons(Cons(Nil, Nil), x), y) gives rise to a decreasing loop by considering the right hand sides subterm at position []. The pumping substitution is [ ]. The result substitution is [x / Cons(Cons(Nil, Nil), 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