/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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NO
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination of the given RelTRS could be disproven:

(0) RelTRS
(1) RelTRSLoopFinderProof [COMPLETE, 41 ms]
(2) NO


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

   f(s(x), y) -> f(-(s(x), y), y)

The relative TRS consists of the following S rules:

   +(0, y) -> y
   +(s(x), y) -> s(+(x, y))
   -(x, 0) -> x
   -(0, y) -> 0
   -(s(x), s(y)) -> -(x, y)
   f(x, y) -> f(x, +(x, y))


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

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

f(-(s(x), 0), 0) -> f(s(x), 0) with rule -(x', 0) -> x' at position [0] and matcher [x' / s(x)]

f(s(x), 0) -> f(-(s(x), 0), 0) with rule f(s(x'), y) -> f(-(s(x'), y), y) at position [] and matcher [x' / x, y / 0]

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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(2)
NO