Termination w.r.t. Q proof of TRS/SK902.59.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:

f₃(g₁(x), y, y) -> g₁(f₃(x, x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₃(g₁(x), y, y) -> g₁(f₃(x, 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:

F₃(g₁(x), y, y) -> F₃(x, x, y)

The TRS R consists of the following rules:

f₃(g₁(x), y, y) -> g₁(f₃(x, 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:

F₃(g₁(x), y, y) -> F₃(x, x, y)

The TRS R consists of the following rules:

f₃(g₁(x), y, y) -> g₁(f₃(x, 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.

F₃(g₁(x), y, y) -> F₃(x, x, y)

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

POL(F₃(x₁, x₂, x₃)) = 2·x₁ + 3·x₁·x₂ + 3·x₁·x₂·x₃ + 3·x₁·x₃ + 3·x₂ + x₂·x₃
POL(g₁(x₁)) = 1 + 3·x₁ + 3·x₁²

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:

f₃(g₁(x), y, y) -> g₁(f₃(x, 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.