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Runti Compl Full Rewri 10127 pair #381902112
details
property
value
status
complete
benchmark
otto06.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n072.star.cs.uiowa.edu
space
AProVE_07
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
291.473562002 seconds
cpu usage
312.765635964
max memory
5.468782592E9
stage attributes
key
value
output-size
5324
starexec-result
WORST_CASE(Omega(n^1), ?)
output
/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^1), ?) proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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: app(x, y) -> helpa(0, plus(length(x), length(y)), x, y) plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) length(nil) -> 0 length(cons(x, y)) -> s(length(y)) helpa(c, l, ys, zs) -> if(ge(c, l), c, l, ys, zs) ge(x, 0) -> true ge(0, s(x)) -> false ge(s(x), s(y)) -> ge(x, y) if(true, c, l, ys, zs) -> nil if(false, c, l, ys, zs) -> helpb(c, l, greater(ys, zs), smaller(ys, zs)) greater(ys, zs) -> helpc(ge(length(ys), length(zs)), ys, zs) smaller(ys, zs) -> helpc(ge(length(ys), length(zs)), zs, ys) helpc(true, ys, zs) -> ys helpc(false, ys, zs) -> zs helpb(c, l, cons(y, ys), zs) -> cons(y, helpa(s(c), l, ys, zs)) 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: app(x, y) -> helpa(0, plus(length(x), length(y)), x, y) plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) length(nil) -> 0 length(cons(x, y)) -> s(length(y)) helpa(c, l, ys, zs) -> if(ge(c, l), c, l, ys, zs) ge(x, 0) -> true ge(0, s(x)) -> false ge(s(x), s(y)) -> ge(x, y) if(true, c, l, ys, zs) -> nil if(false, c, l, ys, zs) -> helpb(c, l, greater(ys, zs), smaller(ys, zs)) greater(ys, zs) -> helpc(ge(length(ys), length(zs)), ys, zs) smaller(ys, zs) -> helpc(ge(length(ys), length(zs)), zs, ys) helpc(true, ys, zs) -> ys helpc(false, ys, zs) -> zs helpb(c, l, cons(y, ys), zs) -> cons(y, helpa(s(c), l, ys, zs)) 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 plus(x, s(y)) ->^+ s(plus(x, y)) gives rise to a decreasing loop by considering the right hand sides subterm at position [0]. The pumping substitution is [y / s(y)]. The result substitution is [ ].
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