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(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(x)

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(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.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(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.

(5) NonTerminationProof (EQUIVALENT transformation)

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:

• Semiunifier: [y / g(g(x''))]
• Matcher: [x' / x'', x'' / 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.

(6) FALSE