details

property	value
status	complete
benchmark	aprove09.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n144.star.cs.uiowa.edu
space	Secret_07_TRS

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	291.516 seconds
cpu usage	315.882
user time	313.625
system time	2.25621
max virtual memory	1.8281672E7
max residence set size	5389372.0

stage attributes

key	value
starexec-result	WORST_CASE(Omega(n^1), ?)

output

WORST_CASE(Omega(n^1), ?) proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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 ---------------------------------------- (0) Obligation: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF). The TRS R consists of the following rules: div(x, y) -> div2(x, y, 0) div2(x, y, i) -> if1(le(y, 0), le(y, x), x, y, plus(i, 0), inc(i)) if1(true, b, x, y, i, j) -> divZeroError if1(false, b, x, y, i, j) -> if2(b, x, y, i, j) if2(true, x, y, i, j) -> div2(minus(x, y), y, j) if2(false, x, y, i, j) -> i inc(0) -> 0 inc(s(i)) -> s(inc(i)) le(s(x), 0) -> false le(0, y) -> true le(s(x), s(y)) -> le(x, y) minus(x, 0) -> x minus(0, y) -> 0 minus(s(x), s(y)) -> minus(x, y) plus(x, y) -> plusIter(x, y, 0) plusIter(x, y, z) -> ifPlus(le(x, z), x, y, z) ifPlus(true, x, y, z) -> y ifPlus(false, x, y, z) -> plusIter(x, s(y), s(z)) a -> c a -> d 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(n^1, INF). The TRS R consists of the following rules: div(x, y) -> div2(x, y, 0) div2(x, y, i) -> if1(le(y, 0), le(y, x), x, y, plus(i, 0), inc(i)) if1(true, b, x, y, i, j) -> divZeroError if1(false, b, x, y, i, j) -> if2(b, x, y, i, j) if2(true, x, y, i, j) -> div2(minus(x, y), y, j) if2(false, x, y, i, j) -> i inc(0) -> 0 inc(s(i)) -> s(inc(i)) le(s(x), 0) -> false le(0, y) -> true le(s(x), s(y)) -> le(x, y) minus(x, 0) -> x minus(0, y) -> 0 minus(s(x), s(y)) -> minus(x, y) plus(x, y) -> plusIter(x, y, 0) plusIter(x, y, z) -> ifPlus(le(x, z), x, y, z) ifPlus(true, x, y, z) -> y ifPlus(false, x, y, z) -> plusIter(x, s(y), s(z)) a -> c a -> d S is empty. Rewrite Strategy: FULL ---------------------------------------- (3) DecreasingLoopProof (LOWER BOUND(ID)) The following loop(s) give(s) rise to the lower bound Omega(n^1): The rewrite sequence le(s(x), s(y)) ->^+ le(x, y) gives rise to a decreasing loop by considering the right hand sides subterm at position []. The pumping substitution is [x / s(x), y / s(y)]. popout

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actions

all output return to Runtime Complexity: TRS