/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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) TransformationProof [EQUIVALENT, 0 ms]
(8) QDP
(9) MNOCProof [EQUIVALENT, 0 ms]
(10) QDP
(11) NonTerminationLoopProof [COMPLETE, 76 ms]
(12) NO


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

   f(f(f(a, x), y), z) -> f(f(x, z), f(y, z))
   f(f(b, x), y) -> x
   f(c, y) -> y

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(f(f(a, x), y), z) -> f(f(x, z), f(y, z))
   f(f(b, x), y) -> x
   f(c, y) -> y

The set Q consists of the following terms:

   f(f(f(a, x0), x1), x2)
   f(f(b, x0), x1)
   f(c, x0)


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

The TRS R consists of the following rules:

   f(f(f(a, x), y), z) -> f(f(x, z), f(y, z))
   f(f(b, x), y) -> x
   f(c, y) -> y

The set Q consists of the following terms:

   f(f(f(a, x0), x1), x2)
   f(f(b, x0), x1)
   f(c, x0)

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

   (F(f(f(a, f(f(a, y_0), y_1)), x1), x2) -> F(f(f(a, y_0), y_1), x2),F(f(f(a, f(f(a, y_0), y_1)), x1), x2) -> F(f(f(a, y_0), y_1), x2))


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

   F(f(f(a, x), y), z) -> F(f(x, z), f(y, z))
   F(f(f(a, x), y), z) -> F(y, z)
   F(f(f(a, f(f(a, y_0), y_1)), x1), x2) -> F(f(f(a, y_0), y_1), x2)

The TRS R consists of the following rules:

   f(f(f(a, x), y), z) -> f(f(x, z), f(y, z))
   f(f(b, x), y) -> x
   f(c, y) -> y

The set Q consists of the following terms:

   f(f(f(a, x0), x1), x2)
   f(f(b, x0), x1)
   f(c, x0)

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

   (F(f(f(a, x0), f(f(a, y_0), y_1)), x2) -> F(f(f(a, y_0), y_1), x2),F(f(f(a, x0), f(f(a, y_0), y_1)), x2) -> F(f(f(a, y_0), y_1), x2))
   (F(f(f(a, x0), f(f(a, f(f(a, y_0), y_1)), y_2)), x2) -> F(f(f(a, f(f(a, y_0), y_1)), y_2), x2),F(f(f(a, x0), f(f(a, f(f(a, y_0), y_1)), y_2)), x2) -> F(f(f(a, f(f(a, y_0), y_1)), y_2), x2))


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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(f(f(a, x), y), z) -> F(f(x, z), f(y, z))
   F(f(f(a, f(f(a, y_0), y_1)), x1), x2) -> F(f(f(a, y_0), y_1), x2)
   F(f(f(a, x0), f(f(a, y_0), y_1)), x2) -> F(f(f(a, y_0), y_1), x2)
   F(f(f(a, x0), f(f(a, f(f(a, y_0), y_1)), y_2)), x2) -> F(f(f(a, f(f(a, y_0), y_1)), y_2), x2)

The TRS R consists of the following rules:

   f(f(f(a, x), y), z) -> f(f(x, z), f(y, z))
   f(f(b, x), y) -> x
   f(c, y) -> y

The set Q consists of the following terms:

   f(f(f(a, x0), x1), x2)
   f(f(b, x0), x1)
   f(c, x0)

We have to consider all minimal (P,Q,R)-chains.
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(9) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
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(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(f(f(a, x), y), z) -> F(f(x, z), f(y, z))
   F(f(f(a, f(f(a, y_0), y_1)), x1), x2) -> F(f(f(a, y_0), y_1), x2)
   F(f(f(a, x0), f(f(a, y_0), y_1)), x2) -> F(f(f(a, y_0), y_1), x2)
   F(f(f(a, x0), f(f(a, f(f(a, y_0), y_1)), y_2)), x2) -> F(f(f(a, f(f(a, y_0), y_1)), y_2), x2)

The TRS R consists of the following rules:

   f(f(f(a, x), y), z) -> f(f(x, z), f(y, z))
   f(f(b, x), y) -> x
   f(c, y) -> y

Q is empty.
We have to consider all (P,Q,R)-chains.
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(11) 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(a, f(a, x0)), y), f(c, f(f(a, y_0), y_1))) evaluates to  t =F(f(f(a, y_0), y_1), f(y, f(f(a, y_0), y_1)))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [ ]
* Semiunifier: [y / c, y_0 / f(a, x0), y_1 / c]

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Rewriting sequence

F(f(f(a, f(a, x0)), c), f(c, f(f(a, f(a, x0)), c))) -> F(f(f(a, f(a, x0)), c), f(f(a, f(a, x0)), c))
with rule f(c, y') -> y' at position [1] and matcher [y' / f(f(a, f(a, x0)), c)]

F(f(f(a, f(a, x0)), c), f(f(a, f(a, x0)), c)) -> F(f(f(a, x0), f(f(a, f(a, x0)), c)), f(c, f(f(a, f(a, x0)), c)))
with rule F(f(f(a, x), y), z) -> F(f(x, z), f(y, z)) at position [] and matcher [x / f(a, x0), y / c, z / f(f(a, f(a, x0)), c)]

F(f(f(a, x0), f(f(a, f(a, x0)), c)), f(c, f(f(a, f(a, x0)), c))) -> F(f(f(a, f(a, x0)), c), f(c, f(f(a, f(a, x0)), c)))
with rule F(f(f(a, x0), f(f(a, y_0), y_1)), x2) -> F(f(f(a, y_0), y_1), x2)

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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(12)
NO