Termination w.r.t. Q proof of /tmp/tmpFHgko5/Ex4_4_Luc96b

Termination w.r.t. Q of the following Term Rewriting System could be disproven:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(g(X), Y) → f(X, n__f(g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
activate(n__f(X1, X2)) → f(X1, X2)
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f(g(X), Y) → f(X, n__f(g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
activate(n__f(X1, X2)) → f(X1, X2)
activate(X) → X

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__f(X1, X2)) → F(X1, X2)
F(g(X), Y) → ACTIVATE(Y)
F(g(X), Y) → F(X, n__f(g(X), activate(Y)))

The TRS R consists of the following rules:

f(g(X), Y) → f(X, n__f(g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
activate(n__f(X1, X2)) → f(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ NonTerminationProof

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__f(X1, X2)) → F(X1, X2)
F(g(X), Y) → ACTIVATE(Y)
F(g(X), Y) → F(X, n__f(g(X), activate(Y)))

The TRS R consists of the following rules:

f(g(X), Y) → f(X, n__f(g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
activate(n__f(X1, X2)) → f(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

ACTIVATE(n__f(X1, X2)) → F(X1, X2)
F(g(X), Y) → ACTIVATE(Y)
F(g(X), Y) → F(X, n__f(g(X), activate(Y)))

The TRS R consists of the following rules:

f(g(X), Y) → f(X, n__f(g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
activate(n__f(X1, X2)) → f(X1, X2)
activate(X) → X

s = F(g(g(X)), Y) evaluates to t =F(g(g(X)), activate(Y))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [Y / activate(Y)]

Rewriting sequence

F(g(g(X)), Y) → F(g(X), n__f(g(g(X)), activate(Y)))
with rule F(g(X'), Y') → F(X', n__f(g(X'), activate(Y'))) at position [] and matcher [Y' / Y, X' / g(X)]

F(g(X), n__f(g(g(X)), activate(Y))) → ACTIVATE(n__f(g(g(X)), activate(Y)))
with rule F(g(X'), Y') → ACTIVATE(Y') at position [] and matcher [Y' / n__f(g(g(X)), activate(Y)), X' / X]

ACTIVATE(n__f(g(g(X)), activate(Y))) → F(g(g(X)), activate(Y))
with rule ACTIVATE(n__f(X1, X2)) → F(X1, 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.