details

property	value
status	complete
benchmark	minsort.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	AProVE_09_Inductive

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	291.514 seconds
cpu usage	306.171
user time	304.354
system time	1.817
max virtual memory	1.8281456E7
max residence set size	5167532.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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: le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) eq(0, 0) -> true eq(0, s(y)) -> false eq(s(x), 0) -> false eq(s(x), s(y)) -> eq(x, y) minsort(nil) -> nil minsort(cons(x, xs)) -> cons(min(cons(x, xs)), minsort(rm(min(cons(x, xs)), cons(x, xs)))) min(nil) -> 0 min(cons(x, nil)) -> x min(cons(x, cons(y, xs))) -> if1(le(x, y), x, y, xs) if1(true, x, y, xs) -> min(cons(x, xs)) if1(false, x, y, xs) -> min(cons(y, xs)) rm(x, nil) -> nil rm(x, cons(y, xs)) -> if2(eq(x, y), x, y, xs) if2(true, x, y, xs) -> rm(x, xs) if2(false, x, y, xs) -> cons(y, rm(x, xs)) 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: le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) eq(0, 0) -> true eq(0, s(y)) -> false eq(s(x), 0) -> false eq(s(x), s(y)) -> eq(x, y) minsort(nil) -> nil minsort(cons(x, xs)) -> cons(min(cons(x, xs)), minsort(rm(min(cons(x, xs)), cons(x, xs)))) min(nil) -> 0 min(cons(x, nil)) -> x min(cons(x, cons(y, xs))) -> if1(le(x, y), x, y, xs) if1(true, x, y, xs) -> min(cons(x, xs)) if1(false, x, y, xs) -> min(cons(y, xs)) rm(x, nil) -> nil rm(x, cons(y, xs)) -> if2(eq(x, y), x, y, xs) if2(true, x, y, xs) -> rm(x, xs) if2(false, x, y, xs) -> cons(y, rm(x, xs)) 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)]. The result substitution is [ ]. popout

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all output return to Runtime Complexity: TRS