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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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

   purge(nil) -> nil
   purge(.(x, y)) -> .(x, purge(remove(x, y)))
   remove(x, nil) -> nil
   remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, 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:

   purge(nil) -> nil
   purge(.(x, y)) -> .(x, purge(remove(x, y)))
   remove(x, nil) -> nil
   remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))

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

remove(x, .(y, z)) ->^+ if(=(x, y), remove(x, z), .(y, remove(x, z)))

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

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

The result substitution is [ ].



The rewrite sequence

remove(x, .(y, z)) ->^+ if(=(x, y), remove(x, z), .(y, remove(x, z)))

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

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

The result substitution is [ ].




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(4)
BOUNDS(EXP, INF)