/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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:

   from(X) -> cons(X, from(s(X)))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)
   minus(X, 0) -> 0
   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)))
   zWquot(XS, nil) -> nil
   zWquot(nil, XS) -> nil
   zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))

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:

   from(X) -> cons(X, from(s(X)))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)
   minus(X, 0) -> 0
   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)))
   zWquot(XS, nil) -> nil
   zWquot(nil, XS) -> nil
   zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))

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

The rewrite sequence

from(X) ->^+ cons(X, from(s(X)))

gives rise to a decreasing loop by considering the right hand sides subterm at position [1].

The pumping substitution is [ ].

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




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(4)
BOUNDS(INF, INF)