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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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

   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, nil)) -> cons(x, nil)
   sum(cons(x, cons(y, l))) -> sum(cons(a(x, y, h), l))
   a(h, h, x) -> s(x)
   a(x, s(y), h) -> a(x, y, s(h))
   a(x, s(y), s(z)) -> a(x, y, a(x, s(y), z))
   a(s(x), h, z) -> a(x, z, z)

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:

   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, nil)) -> cons(x, nil)
   sum(cons(x, cons(y, l))) -> sum(cons(a(x, y, h), l))
   a(h, h, x) -> s(x)
   a(x, s(y), h) -> a(x, y, s(h))
   a(x, s(y), s(z)) -> a(x, y, a(x, s(y), z))
   a(s(x), h, z) -> a(x, z, z)

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

app(cons(x, l), k) ->^+ cons(x, app(l, k))

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

The pumping substitution is [l / cons(x, l)].

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(EXP, INF).


The TRS R consists of the following rules:

   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, nil)) -> cons(x, nil)
   sum(cons(x, cons(y, l))) -> sum(cons(a(x, y, h), l))
   a(h, h, x) -> s(x)
   a(x, s(y), h) -> a(x, y, s(h))
   a(x, s(y), s(z)) -> a(x, y, a(x, s(y), z))
   a(s(x), h, z) -> a(x, z, z)

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:

   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, nil)) -> cons(x, nil)
   sum(cons(x, cons(y, l))) -> sum(cons(a(x, y, h), l))
   a(h, h, x) -> s(x)
   a(x, s(y), h) -> a(x, y, s(h))
   a(x, s(y), s(z)) -> a(x, y, a(x, s(y), z))
   a(s(x), h, z) -> a(x, z, z)

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

a(s(x1_0), s(h), s(z)) ->^+ a(x1_0, a(s(x1_0), s(h), z), a(s(x1_0), s(h), z))

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

The pumping substitution is [z / s(z)].

The result substitution is [ ].



The rewrite sequence

a(s(x1_0), s(h), s(z)) ->^+ a(x1_0, a(s(x1_0), s(h), z), a(s(x1_0), s(h), z))

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

The pumping substitution is [z / s(z)].

The result substitution is [ ].




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(10)
BOUNDS(EXP, INF)