Termination w.r.t. Q proof of TRS/nontermin/Rubio-inntest75.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f₃(0, 1, X) -> f₃(g₂(X, X), X, X)
g₂(X, Y) -> X
g₂(X, Y) -> Y

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₃(0, 1, X) -> f₃(g₂(X, X), X, X)
g₂(X, Y) -> X
g₂(X, Y) -> Y

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F₃(0, 1, X) -> F₃(g₂(X, X), X, X)
F₃(0, 1, X) -> G₂(X, X)

The TRS R consists of the following rules:

f₃(0, 1, X) -> f₃(g₂(X, X), X, X)
g₂(X, Y) -> X
g₂(X, Y) -> Y

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

F₃(0, 1, X) -> F₃(g₂(X, X), X, X)
F₃(0, 1, X) -> G₂(X, X)

The TRS R consists of the following rules:

f₃(0, 1, X) -> f₃(g₂(X, X), X, X)
g₂(X, Y) -> X
g₂(X, Y) -> Y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

F₃(0, 1, X) -> F₃(g₂(X, X), X, X)

The TRS R consists of the following rules:

f₃(0, 1, X) -> f₃(g₂(X, X), X, X)
g₂(X, Y) -> X
g₂(X, Y) -> Y

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