/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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WORST_CASE(NON_POLY, ?)
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).

(0) CpxTRS
(1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
(2) TRS for Loop Detection
(3) InfiniteLowerBoundProof [FINISHED, 0 ms]
(4) BOUNDS(INF, INF)


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(0)
Obligation:
The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).


The TRS R consists of the following rules:

   p(0) -> 0
   p(s(X)) -> X
   leq(0, Y) -> true
   leq(s(X), 0) -> false
   leq(s(X), s(Y)) -> leq(X, Y)
   if(true, X, Y) -> activate(X)
   if(false, X, Y) -> activate(Y)
   diff(X, Y) -> if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
   0 -> n__0
   s(X) -> n__s(X)
   activate(n__0) -> 0
   activate(n__s(X)) -> s(X)
   activate(X) -> X

S is empty.
Rewrite Strategy: FULL
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(1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
Transformed a relative TRS into a decreasing-loop problem.
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(2)
Obligation:
Analyzing the following TRS for decreasing loops:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).


The TRS R consists of the following rules:

   p(0) -> 0
   p(s(X)) -> X
   leq(0, Y) -> true
   leq(s(X), 0) -> false
   leq(s(X), s(Y)) -> leq(X, Y)
   if(true, X, Y) -> activate(X)
   if(false, X, Y) -> activate(Y)
   diff(X, Y) -> if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
   0 -> n__0
   s(X) -> n__s(X)
   activate(n__0) -> 0
   activate(n__s(X)) -> s(X)
   activate(X) -> X

S is empty.
Rewrite Strategy: FULL
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(3) InfiniteLowerBoundProof (FINISHED)
The following loop proves infinite runtime complexity:

The rewrite sequence

diff(X, Y) ->^+ if(leq(X, Y), n__0, n__s(diff(p(X), Y)))

gives rise to a decreasing loop by considering the right hand sides subterm at position [2,0].

The pumping substitution is [ ].

The result substitution is [X / p(X)].




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(4)
BOUNDS(INF, INF)