/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(EXP, 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) DecreasingLoopProof [FINISHED, 0 ms]
        (10) BOUNDS(EXP, INF)


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


The TRS R consists of the following rules:

   rev1(0, nil) -> 0
   rev1(s(X), nil) -> s(X)
   rev1(X, cons(Y, L)) -> rev1(Y, L)
   rev(nil) -> nil
   rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
   rev2(X, nil) -> nil
   rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

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(EXP, INF).


The TRS R consists of the following rules:

   rev1(0, nil) -> 0
   rev1(s(X), nil) -> s(X)
   rev1(X, cons(Y, L)) -> rev1(Y, L)
   rev(nil) -> nil
   rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
   rev2(X, nil) -> nil
   rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

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

rev2(X, cons(Y, L)) ->^+ rev(cons(X, rev(rev2(Y, L))))

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

The pumping substitution is [L / cons(Y, L)].

The result substitution is [X / Y].




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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(EXP, INF).


The TRS R consists of the following rules:

   rev1(0, nil) -> 0
   rev1(s(X), nil) -> s(X)
   rev1(X, cons(Y, L)) -> rev1(Y, L)
   rev(nil) -> nil
   rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
   rev2(X, nil) -> nil
   rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

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(EXP, INF).


The TRS R consists of the following rules:

   rev1(0, nil) -> 0
   rev1(s(X), nil) -> s(X)
   rev1(X, cons(Y, L)) -> rev1(Y, L)
   rev(nil) -> nil
   rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
   rev2(X, nil) -> nil
   rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

S is empty.
Rewrite Strategy: FULL
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(9) DecreasingLoopProof (FINISHED)
The following loop(s) give(s) rise to the lower bound EXP:

The rewrite sequence

rev2(X, cons(Y, L)) ->^+ cons(rev1(X, rev(rev2(Y, L))), rev2(X, rev(rev2(Y, L))))

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

The pumping substitution is [L / cons(Y, L)].

The result substitution is [X / Y].



The rewrite sequence

rev2(X, cons(Y, L)) ->^+ cons(rev1(X, rev(rev2(Y, L))), rev2(X, rev(rev2(Y, L))))

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

The pumping substitution is [L / cons(Y, L)].

The result substitution is [X / Y].




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(10)
BOUNDS(EXP, INF)