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Runtime Complexity: TRS Innermost pair #487112574
details
property
value
status
complete
benchmark
aprove05.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n142.star.cs.uiowa.edu
space
Secret_07_TRS
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
292.152 seconds
cpu usage
1100.79
user time
1087.2
system time
13.5966
max virtual memory
3.7782048E7
max residence set size
1.4917548E7
stage attributes
key
value
starexec-result
WORST_CASE(Omega(n^1), ?)
output
WORST_CASE(Omega(n^1), ?) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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: prod(xs) -> prodIter(xs, s(0)) prodIter(xs, x) -> ifProd(isempty(xs), xs, x) ifProd(true, xs, x) -> x ifProd(false, xs, x) -> prodIter(tail(xs), times(x, head(xs))) plus(0, y) -> y plus(s(x), y) -> s(plus(x, y)) times(x, y) -> timesIter(x, y, 0, 0) timesIter(x, y, z, u) -> ifTimes(ge(u, x), x, y, z, u) ifTimes(true, x, y, z, u) -> z ifTimes(false, x, y, z, u) -> timesIter(x, y, plus(y, z), s(u)) isempty(nil) -> true isempty(cons(x, xs)) -> false head(nil) -> error head(cons(x, xs)) -> x tail(nil) -> nil tail(cons(x, xs)) -> xs ge(x, 0) -> true ge(0, s(y)) -> false ge(s(x), s(y)) -> ge(x, y) a -> b a -> c 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: prod(xs) -> prodIter(xs, s(0)) prodIter(xs, x) -> ifProd(isempty(xs), xs, x) ifProd(true, xs, x) -> x ifProd(false, xs, x) -> prodIter(tail(xs), times(x, head(xs))) plus(0, y) -> y plus(s(x), y) -> s(plus(x, y)) times(x, y) -> timesIter(x, y, 0, 0) timesIter(x, y, z, u) -> ifTimes(ge(u, x), x, y, z, u) ifTimes(true, x, y, z, u) -> z ifTimes(false, x, y, z, u) -> timesIter(x, y, plus(y, z), s(u)) isempty(nil) -> true isempty(cons(x, xs)) -> false head(nil) -> error head(cons(x, xs)) -> x tail(nil) -> nil tail(cons(x, xs)) -> xs ge(x, 0) -> true ge(0, s(y)) -> false ge(s(x), s(y)) -> ge(x, y) a -> b a -> c 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 plus(s(x), y) ->^+ s(plus(x, y)) gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
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