Termination w.r.t. Q proof of TRS/SK902.07.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(0, y) -> y
f₂(x, 0) -> x
f₂(i₁(x), y) -> i₁(x)
f₂(f₂(x, y), z) -> f₂(x, f₂(y, z))
f₂(g₂(x, y), z) -> g₂(f₂(x, z), f₂(y, z))
f₂(1, g₂(x, y)) -> x
f₂(2, g₂(x, y)) -> y

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(0, y) -> y
f₂(x, 0) -> x
f₂(i₁(x), y) -> i₁(x)
f₂(f₂(x, y), z) -> f₂(x, f₂(y, z))
f₂(g₂(x, y), z) -> g₂(f₂(x, z), f₂(y, z))
f₂(1, g₂(x, y)) -> x
f₂(2, 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:

F₂(g₂(x, y), z) -> F₂(y, z)
F₂(g₂(x, y), z) -> F₂(x, z)
F₂(f₂(x, y), z) -> F₂(x, f₂(y, z))
F₂(f₂(x, y), z) -> F₂(y, z)

The TRS R consists of the following rules:

f₂(0, y) -> y
f₂(x, 0) -> x
f₂(i₁(x), y) -> i₁(x)
f₂(f₂(x, y), z) -> f₂(x, f₂(y, z))
f₂(g₂(x, y), z) -> g₂(f₂(x, z), f₂(y, z))
f₂(1, g₂(x, y)) -> x
f₂(2, g₂(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:

F₂(g₂(x, y), z) -> F₂(y, z)
F₂(g₂(x, y), z) -> F₂(x, z)
F₂(f₂(x, y), z) -> F₂(x, f₂(y, z))
F₂(f₂(x, y), z) -> F₂(y, z)

The TRS R consists of the following rules:

f₂(0, y) -> y
f₂(x, 0) -> x
f₂(i₁(x), y) -> i₁(x)
f₂(f₂(x, y), z) -> f₂(x, f₂(y, z))
f₂(g₂(x, y), z) -> g₂(f₂(x, z), f₂(y, z))
f₂(1, g₂(x, y)) -> x
f₂(2, g₂(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.

F₂(g₂(x, y), z) -> F₂(y, z)
F₂(g₂(x, y), z) -> F₂(x, z)

The remaining pairs can at least be oriented weakly.

F₂(f₂(x, y), z) -> F₂(x, f₂(y, z))
F₂(f₂(x, y), z) -> F₂(y, z)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(1) = 0
POL(2) = 0
POL(F₂(x₁, x₂)) = x₁
POL(f₂(x₁, x₂)) = 2·x₁ + x₂
POL(g₂(x₁, x₂)) = 3 + x₁ + 2·x₂
POL(i₁(x₁)) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

F₂(f₂(x, y), z) -> F₂(x, f₂(y, z))
F₂(f₂(x, y), z) -> F₂(y, z)

The TRS R consists of the following rules:

f₂(0, y) -> y
f₂(x, 0) -> x
f₂(i₁(x), y) -> i₁(x)
f₂(f₂(x, y), z) -> f₂(x, f₂(y, z))
f₂(g₂(x, y), z) -> g₂(f₂(x, z), f₂(y, z))
f₂(1, g₂(x, y)) -> x
f₂(2, g₂(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.

F₂(f₂(x, y), z) -> F₂(x, f₂(y, z))
F₂(f₂(x, y), z) -> F₂(y, z)

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

POL(0) = 0
POL(1) = 0
POL(2) = 0
POL(F₂(x₁, x₂)) = 3·x₁
POL(f₂(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(g₂(x₁, x₂)) = 0
POL(i₁(x₁)) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ PisEmptyProof

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

f₂(0, y) -> y
f₂(x, 0) -> x
f₂(i₁(x), y) -> i₁(x)
f₂(f₂(x, y), z) -> f₂(x, f₂(y, z))
f₂(g₂(x, y), z) -> g₂(f₂(x, z), f₂(y, z))
f₂(1, g₂(x, y)) -> x
f₂(2, g₂(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.