/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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NO
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination of the given RelTRS could be disproven:

(0) RelTRS
(1) RelTRS Reverse [EQUIVALENT, 0 ms]
(2) RelTRS
(3) RelTRSLoopFinderProof [COMPLETE, 3 ms]
(4) NO


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(0)
Obligation:
Relative term rewrite system:
The relative TRS consists of the following R rules:

   a(b(c(x1))) -> c(x1)

The relative TRS consists of the following S rules:

   c(x1) -> c(b(a(x1)))


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(1) RelTRS Reverse (EQUIVALENT)
We have reversed the following relative TRS [REVERSE]:
The set of rules R is 
   a(b(c(x1))) -> c(x1)

The set of rules S is 
   c(x1) -> c(b(a(x1)))

We have obtained the following relative TRS:
The set of rules R is 
   c(b(a(x1))) -> c(x1)

The set of rules S is 
   c(x1) -> a(b(c(x1)))


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(2)
Obligation:
Relative term rewrite system:
The relative TRS consists of the following R rules:

   c(b(a(x1))) -> c(x1)

The relative TRS consists of the following S rules:

   c(x1) -> a(b(c(x1)))


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(3) RelTRSLoopFinderProof (COMPLETE)
The following loop was found:

---------- Loop: ----------

c(b(c(x1'))) -> c(b(a(b(c(x1'))))) with rule c(x1'') -> a(b(c(x1''))) at position [0,0] and matcher [x1'' / x1']

c(b(a(b(c(x1'))))) -> c(b(c(x1'))) with rule c(b(a(x1))) -> c(x1) at position [] and matcher [x1 / b(c(x1'))]

Now an instance of the first term with Matcher [ ] occurs in the last term at position [].

Context: []

Therefore, the relative TRS problem does not terminate.
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(4)
NO