/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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WORST_CASE(NON_POLY, ?)
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) 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
        (9) InfiniteLowerBoundProof [FINISHED, 71 ms]
        (10) 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:

   nats -> cons(0, n__incr(n__nats))
   pairs -> cons(0, n__incr(n__odds))
   odds -> incr(pairs)
   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
   head(cons(X, XS)) -> X
   tail(cons(X, XS)) -> activate(XS)
   incr(X) -> n__incr(X)
   nats -> n__nats
   odds -> n__odds
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__nats) -> nats
   activate(n__odds) -> odds
   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:

   nats -> cons(0, n__incr(n__nats))
   pairs -> cons(0, n__incr(n__odds))
   odds -> incr(pairs)
   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
   head(cons(X, XS)) -> X
   tail(cons(X, XS)) -> activate(XS)
   incr(X) -> n__incr(X)
   nats -> n__nats
   odds -> n__odds
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__nats) -> nats
   activate(n__odds) -> odds
   activate(X) -> X

S is empty.
Rewrite Strategy: FULL
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(3) DecreasingLoopProof (LOWER BOUND(ID))
The following loop(s) give(s) rise to the lower bound Omega(n^1):

The rewrite sequence

activate(n__incr(X)) ->^+ incr(activate(X))

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

The pumping substitution is [X / n__incr(X)].

The result substitution is [ ].




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(4)
Complex Obligation (BEST)

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(5)
Obligation:
Proved the lower bound n^1 for the following 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:

   nats -> cons(0, n__incr(n__nats))
   pairs -> cons(0, n__incr(n__odds))
   odds -> incr(pairs)
   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
   head(cons(X, XS)) -> X
   tail(cons(X, XS)) -> activate(XS)
   incr(X) -> n__incr(X)
   nats -> n__nats
   odds -> n__odds
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__nats) -> nats
   activate(n__odds) -> odds
   activate(X) -> X

S is empty.
Rewrite Strategy: FULL
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(6) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
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(7)
BOUNDS(n^1, INF)

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(8)
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:

   nats -> cons(0, n__incr(n__nats))
   pairs -> cons(0, n__incr(n__odds))
   odds -> incr(pairs)
   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
   head(cons(X, XS)) -> X
   tail(cons(X, XS)) -> activate(XS)
   incr(X) -> n__incr(X)
   nats -> n__nats
   odds -> n__odds
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__nats) -> nats
   activate(n__odds) -> odds
   activate(X) -> X

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

The rewrite sequence

incr(cons(X, n__incr(n__nats))) ->^+ cons(s(X), n__incr(incr(cons(0, n__incr(n__nats)))))

gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0].

The pumping substitution is [ ].

The result substitution is [X / 0].




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(10)
BOUNDS(INF, INF)