NO
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Termination w.r.t. Q of the given QTRS could be disproven:

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


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

   f(x, y, f(z, u, v)) -> f(f(x, y, z), u, f(x, y, v))

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(x, y, f(z, u, v)) -> F(f(x, y, z), u, f(x, y, v))
   F(x, y, f(z, u, v)) -> F(x, y, z)
   F(x, y, f(z, u, v)) -> F(x, y, v)

The TRS R consists of the following rules:

   f(x, y, f(z, u, v)) -> f(f(x, y, z), u, f(x, y, v))

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

s = F(x, y, f(z, u, v)) evaluates to  t =F(f(x, y, z), u, f(x, y, v))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [x / f(x, y, z), y / u, z / x, u / y]
* Semiunifier: [ ]

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from F(x, y, f(z, u, v)) to F(f(x, y, z), u, f(x, y, v)).




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(4)
NO