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

(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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(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.

(3) NonTerminationProof (EQUIVALENT transformation)

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)).

(4) NO