/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(EXP, INF).

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

   merge(nil, y) -> y
   merge(x, nil) -> x
   merge(.(x, y), .(u, v)) -> if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
   ++(nil, y) -> y
   ++(.(x, y), z) -> .(x, ++(y, z))
   if(true, x, y) -> x
   if(false, x, y) -> 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(EXP, INF).


The TRS R consists of the following rules:

   merge(nil, y) -> y
   merge(x, nil) -> x
   merge(.(x, y), .(u, v)) -> if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
   ++(nil, y) -> y
   ++(.(x, y), z) -> .(x, ++(y, z))
   if(true, x, y) -> x
   if(false, x, y) -> x

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

The rewrite sequence

merge(.(x, y), .(u, v)) ->^+ if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))

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

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

The result substitution is [ ].



The rewrite sequence

merge(.(x, y), .(u, v)) ->^+ if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))

gives rise to a decreasing loop by considering the right hand sides subterm at position [2,1].

The pumping substitution is [v / .(u, v)].

The result substitution is [ ].




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(4)
BOUNDS(EXP, INF)