details

property	value
status	complete
benchmark	otto10.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n057.star.cs.uiowa.edu
space	AProVE_07

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	291.898926973 seconds
cpu usage	1150.95132214
max memory	1.5636250624E10

stage attributes

key	value
output-size	4439
starexec-result	WORST_CASE(Omega(n^1), ?)

output

/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^1), ?) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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 (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, INF). The TRS R consists of the following rules: ge(x, 0) -> true ge(0, s(y)) -> false ge(s(x), s(y)) -> ge(x, y) rev(x) -> if(x, eq(0, length(x)), nil, 0, length(x)) if(x, true, z, c, l) -> z if(x, false, z, c, l) -> help(s(c), l, x, z) help(c, l, cons(x, y), z) -> if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l) append(nil, y) -> y append(cons(x, y), z) -> cons(x, append(y, z)) length(nil) -> 0 length(cons(x, y)) -> s(length(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(n^1, INF). The TRS R consists of the following rules: ge(x, 0) -> true ge(0, s(y)) -> false ge(s(x), s(y)) -> ge(x, y) rev(x) -> if(x, eq(0, length(x)), nil, 0, length(x)) if(x, true, z, c, l) -> z if(x, false, z, c, l) -> help(s(c), l, x, z) help(c, l, cons(x, y), z) -> if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l) append(nil, y) -> y append(cons(x, y), z) -> cons(x, append(y, z)) length(nil) -> 0 length(cons(x, y)) -> s(length(y)) S is empty. Rewrite Strategy: INNERMOST ---------------------------------------- (3) DecreasingLoopProof (LOWER BOUND(ID)) The following loop(s) give(s) rise to the lower bound Omega(n^1): The rewrite sequence length(cons(x, y)) ->^+ s(length(y)) gives rise to a decreasing loop by considering the right hand sides subterm at position [0]. The pumping substitution is [y / cons(x, y)]. The result substitution is [ ]. ---------------------------------------- (4) Complex Obligation (BEST) ---------------------------------------- popout

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all output return to Runti Compl Inner Rewri 22807