Termination w.r.t. Q proof of TRS/nontermin/CSREx4_4

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:

f₂(g₁(X), Y) -> f₂(X, f₂(g₁(X), Y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(g₁(X), Y) -> f₂(X, f₂(g₁(X), 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:

F₂(g₁(X), Y) -> F₂(X, f₂(g₁(X), Y))
F₂(g₁(X), Y) -> F₂(g₁(X), Y)

The TRS R consists of the following rules:

f₂(g₁(X), Y) -> f₂(X, f₂(g₁(X), Y))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

F₂(g₁(X), Y) -> F₂(X, f₂(g₁(X), Y))
F₂(g₁(X), Y) -> F₂(g₁(X), Y)

The TRS R consists of the following rules:

f₂(g₁(X), Y) -> f₂(X, f₂(g₁(X), Y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

F₂(g₁(X), Y) -> F₂(X, f₂(g₁(X), Y))

The remaining pairs can at least be oriented weakly.

F₂(g₁(X), Y) -> F₂(g₁(X), Y)

Used ordering: Polynomial interpretation [21]:

POL(F₂(x₁, x₂)) = x₁
POL(f₂(x₁, x₂)) = 0
POL(g₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

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

F₂(g₁(X), Y) -> F₂(g₁(X), Y)

The TRS R consists of the following rules:

f₂(g₁(X), Y) -> f₂(X, f₂(g₁(X), Y))

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