Termination w.r.t. Q proof of TRS/SK902.55.trs

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

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

f₂(x, g₁(x)) -> x
f₂(x, h₁(y)) -> f₂(h₁(x), y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(x, g₁(x)) -> x
f₂(x, h₁(y)) -> f₂(h₁(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(x, h₁(y)) -> F₂(h₁(x), y)

The TRS R consists of the following rules:

f₂(x, g₁(x)) -> x
f₂(x, h₁(y)) -> f₂(h₁(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(x, h₁(y)) -> F₂(h₁(x), y)

The TRS R consists of the following rules:

f₂(x, g₁(x)) -> x
f₂(x, h₁(y)) -> f₂(h₁(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(x, h₁(y)) -> F₂(h₁(x), y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F₂(x₁, x₂)) = 2·x₂
POL(h₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f₂(x, g₁(x)) -> x
f₂(x, h₁(y)) -> f₂(h₁(x), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.