Termination w.r.t. Q proof of TRS/SK904.48.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(f₃(x, y, z), u, f₃(x, y, v)) -> f₃(x, y, f₃(z, u, v))
f₃(x, y, y) -> y
f₃(x, y, g₁(y)) -> x
f₃(x, x, y) -> x
f₃(g₁(x), x, y) -> y

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₃(f₃(x, y, z), u, f₃(x, y, v)) -> f₃(x, y, f₃(z, u, v))
f₃(x, y, y) -> y
f₃(x, y, g₁(y)) -> x
f₃(x, x, y) -> x
f₃(g₁(x), 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(f₃(x, y, z), u, f₃(x, y, v)) -> F₃(x, y, f₃(z, u, v))
F₃(f₃(x, y, z), u, f₃(x, y, v)) -> F₃(z, u, v)

The TRS R consists of the following rules:

f₃(f₃(x, y, z), u, f₃(x, y, v)) -> f₃(x, y, f₃(z, u, v))
f₃(x, y, y) -> y
f₃(x, y, g₁(y)) -> x
f₃(x, x, y) -> x
f₃(g₁(x), x, y) -> 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(f₃(x, y, z), u, f₃(x, y, v)) -> F₃(x, y, f₃(z, u, v))
F₃(f₃(x, y, z), u, f₃(x, y, v)) -> F₃(z, u, v)

The TRS R consists of the following rules:

f₃(f₃(x, y, z), u, f₃(x, y, v)) -> f₃(x, y, f₃(z, u, v))
f₃(x, y, y) -> y
f₃(x, y, g₁(y)) -> x
f₃(x, x, y) -> x
f₃(g₁(x), x, y) -> 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(f₃(x, y, z), u, f₃(x, y, v)) -> F₃(x, y, f₃(z, u, v))
F₃(f₃(x, y, z), u, f₃(x, y, v)) -> F₃(z, u, v)

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

POL(F₃(x₁, x₂, x₃)) = x₁
POL(f₃(x₁, x₂, x₃)) = 1 + x₁ + x₃
POL(g₁(x₁)) = 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:

f₃(f₃(x, y, z), u, f₃(x, y, v)) -> f₃(x, y, f₃(z, u, v))
f₃(x, y, y) -> y
f₃(x, y, g₁(y)) -> x
f₃(x, x, y) -> x
f₃(g₁(x), x, y) -> 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.