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Runti Compl Full Rewri 10127 pair #381902891
details
property
value
status
complete
benchmark
wiehe11.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n054.star.cs.uiowa.edu
space
AProVE_07
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
291.43442297 seconds
cpu usage
346.265454128
max memory
6.230347776E9
stage attributes
key
value
output-size
6240
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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))))) plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x)) plus(zero, y) -> y plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y)))) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y not(x) -> if(x, false, true) gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y) 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))))) plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x)) plus(zero, y) -> y plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y)))) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y not(x) -> if(x, false, true) gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y)
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