/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) TransformationProof [EQUIVALENT, 0 ms]
(6) QDP
(7) MNOCProof [EQUIVALENT, 0 ms]
(8) QDP
(9) NonTerminationLoopProof [COMPLETE, 0 ms]
(10) NO


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

   f(h(x), y) -> h(f(y, f(x, h(a))))

Q is empty.

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(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(h(x), y) -> h(f(y, f(x, h(a))))

The set Q consists of the following terms:

   f(h(x0), x1)


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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(h(x), y) -> F(y, f(x, h(a)))
   F(h(x), y) -> F(x, h(a))

The TRS R consists of the following rules:

   f(h(x), y) -> h(f(y, f(x, h(a))))

The set Q consists of the following terms:

   f(h(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
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(5) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule F(h(x), y) -> F(x, h(a)) we obtained the following new rules [LPAR04]:

   (F(h(h(y_0)), x1) -> F(h(y_0), h(a)),F(h(h(y_0)), x1) -> F(h(y_0), h(a)))


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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(h(x), y) -> F(y, f(x, h(a)))
   F(h(h(y_0)), x1) -> F(h(y_0), h(a))

The TRS R consists of the following rules:

   f(h(x), y) -> h(f(y, f(x, h(a))))

The set Q consists of the following terms:

   f(h(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
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(7) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(h(x), y) -> F(y, f(x, h(a)))
   F(h(h(y_0)), x1) -> F(h(y_0), h(a))

The TRS R consists of the following rules:

   f(h(x), y) -> h(f(y, f(x, h(a))))

Q is empty.
We have to consider all (P,Q,R)-chains.
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(9) 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(f(f(h(x), y''), y'), y) evaluates to  t =F(y, f(f(y', f(f(y'', f(x, h(a))), h(a))), h(a)))

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

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Rewriting sequence

F(f(f(h(x), y''), h(x'')), f(f(h(x'), y'''), h(x'''))) -> F(f(h(f(y'', f(x, h(a)))), h(x'')), f(f(h(x'), y'''), h(x''')))
with rule f(h(x1), y'''') -> h(f(y'''', f(x1, h(a)))) at position [0,0] and matcher [x1 / x, y'''' / y'']

F(f(h(f(y'', f(x, h(a)))), h(x'')), f(f(h(x'), y'''), h(x'''))) -> F(h(f(h(x''), f(f(y'', f(x, h(a))), h(a)))), f(f(h(x'), y'''), h(x''')))
with rule f(h(x'1), y') -> h(f(y', f(x'1, h(a)))) at position [0] and matcher [x'1 / f(y'', f(x, h(a))), y' / h(x'')]

F(h(f(h(x''), f(f(y'', f(x, h(a))), h(a)))), f(f(h(x'), y'''), h(x'''))) -> F(f(f(h(x'), y'''), h(x''')), f(f(h(x''), f(f(y'', f(x, h(a))), h(a))), h(a)))
with rule F(h(x), y) -> F(y, f(x, h(a)))

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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(10)
NO