/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 w.r.t. Q of the given QTRS could be disproven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) NonTerminationLoopProof [COMPLETE, 0 ms]
(6) NO


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

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

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(g(x), x, y) -> F(y, y, g(y))
   F(g(x), x, y) -> G(y)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(g(x), x, y) -> F(y, y, g(y))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = F(g(g(x')), g(g(x')), y) evaluates to  t =F(y, y, g(y))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [x' / x'', x'' / g(x'')]
* Semiunifier: [y / g(g(x''))]

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Rewriting sequence

F(g(g(x')), g(g(x')), g(g(x''))) -> F(g(g(x')), g(x'), g(g(x'')))
with rule g(g(x''')) -> g(x''') at position [1] and matcher [x''' / x']

F(g(g(x')), g(x'), g(g(x''))) -> F(g(g(x'')), g(g(x'')), g(g(g(x''))))
with rule F(g(x), x, y) -> F(y, y, g(y))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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(6)
NO