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Runti Compl Full Rewri 10127 pair #381902706
details
property
value
status
complete
benchmark
Liveness6.4.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n073.star.cs.uiowa.edu
space
AProVE_04
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
293.853926182 seconds
cpu usage
1098.14295925
max memory
1.5554433024E10
stage attributes
key
value
output-size
5169
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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: top1(free(x), y) -> top2(check(new(x)), y) top1(free(x), y) -> top2(new(x), check(y)) top1(free(x), y) -> top2(check(x), new(y)) top1(free(x), y) -> top2(x, check(new(y))) top2(x, free(y)) -> top1(check(new(x)), y) top2(x, free(y)) -> top1(new(x), check(y)) top2(x, free(y)) -> top1(check(x), new(y)) top2(x, free(y)) -> top1(x, check(new(y))) new(free(x)) -> free(new(x)) old(free(x)) -> free(old(x)) new(serve) -> free(serve) old(serve) -> free(serve) check(free(x)) -> free(check(x)) check(new(x)) -> new(check(x)) check(old(x)) -> old(check(x)) check(old(x)) -> old(x) 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: top1(free(x), y) -> top2(check(new(x)), y) top1(free(x), y) -> top2(new(x), check(y)) top1(free(x), y) -> top2(check(x), new(y)) top1(free(x), y) -> top2(x, check(new(y))) top2(x, free(y)) -> top1(check(new(x)), y) top2(x, free(y)) -> top1(new(x), check(y)) top2(x, free(y)) -> top1(check(x), new(y)) top2(x, free(y)) -> top1(x, check(new(y))) new(free(x)) -> free(new(x)) old(free(x)) -> free(old(x)) new(serve) -> free(serve) old(serve) -> free(serve) check(free(x)) -> free(check(x)) check(new(x)) -> new(check(x)) check(old(x)) -> old(check(x)) check(old(x)) -> old(x) 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 check(free(x)) ->^+ free(check(x)) gives rise to a decreasing loop by considering the right hand sides subterm at position [0]. The pumping substitution is [x / free(x)]. The result substitution is [ ].
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