details

property	value
status	complete
benchmark	#3.57.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n052.star.cs.uiowa.edu
space	AG01

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	291.531327009 seconds
cpu usage	894.228863803
max memory	1.1414634496E10

stage attributes

key	value
output-size	4632
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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: minus(x, 0) -> x minus(s(x), s(y)) -> minus(x, y) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) plus(0, y) -> y plus(s(x), y) -> s(plus(x, y)) minus(minus(x, y), z) -> minus(x, plus(y, z)) app(nil, k) -> k app(l, nil) -> l app(cons(x, l), k) -> cons(x, app(l, k)) sum(cons(x, nil)) -> cons(x, nil) sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l)) sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k))))) 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: minus(x, 0) -> x minus(s(x), s(y)) -> minus(x, y) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) plus(0, y) -> y plus(s(x), y) -> s(plus(x, y)) minus(minus(x, y), z) -> minus(x, plus(y, z)) app(nil, k) -> k app(l, nil) -> l app(cons(x, l), k) -> cons(x, app(l, k)) sum(cons(x, nil)) -> cons(x, nil) sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l)) sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k))))) 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 app(cons(x, l), k) ->^+ cons(x, app(l, k)) gives rise to a decreasing loop by considering the right hand sides subterm at position [1]. The pumping substitution is [l / cons(x, l)]. The result substitution is [ ]. ---------------------------------------- popout

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job log

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actions

all output return to Runti Compl Full Rewri 10127