Termination w.r.t. Q proof of TRS/SK904.42.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(a, g₁(y)) -> g₁(g₁(y))
f₂(g₁(x), a) -> f₂(x, g₁(a))
f₂(g₁(x), g₁(y)) -> h₃(g₁(y), x, g₁(y))
h₃(g₁(x), y, z) -> f₂(y, h₃(x, y, z))
h₃(a, y, z) -> z

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(a, g₁(y)) -> g₁(g₁(y))
f₂(g₁(x), a) -> f₂(x, g₁(a))
f₂(g₁(x), g₁(y)) -> h₃(g₁(y), x, g₁(y))
h₃(g₁(x), y, z) -> f₂(y, h₃(x, y, z))
h₃(a, y, z) -> z

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:

H₃(g₁(x), y, z) -> H₃(x, y, z)
H₃(g₁(x), y, z) -> F₂(y, h₃(x, y, z))
F₂(g₁(x), g₁(y)) -> H₃(g₁(y), x, g₁(y))
F₂(g₁(x), a) -> F₂(x, g₁(a))

The TRS R consists of the following rules:

f₂(a, g₁(y)) -> g₁(g₁(y))
f₂(g₁(x), a) -> f₂(x, g₁(a))
f₂(g₁(x), g₁(y)) -> h₃(g₁(y), x, g₁(y))
h₃(g₁(x), y, z) -> f₂(y, h₃(x, y, z))
h₃(a, y, z) -> z

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:

H₃(g₁(x), y, z) -> H₃(x, y, z)
H₃(g₁(x), y, z) -> F₂(y, h₃(x, y, z))
F₂(g₁(x), g₁(y)) -> H₃(g₁(y), x, g₁(y))
F₂(g₁(x), a) -> F₂(x, g₁(a))

The TRS R consists of the following rules:

f₂(a, g₁(y)) -> g₁(g₁(y))
f₂(g₁(x), a) -> f₂(x, g₁(a))
f₂(g₁(x), g₁(y)) -> h₃(g₁(y), x, g₁(y))
h₃(g₁(x), y, z) -> f₂(y, h₃(x, y, z))
h₃(a, y, z) -> z

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.

H₃(g₁(x), y, z) -> F₂(y, h₃(x, y, z))
F₂(g₁(x), g₁(y)) -> H₃(g₁(y), x, g₁(y))
F₂(g₁(x), a) -> F₂(x, g₁(a))

The remaining pairs can at least be oriented weakly.

H₃(g₁(x), y, z) -> H₃(x, y, z)

Used ordering: Polynomial interpretation [21]:

POL(F₂(x₁, x₂)) = 2·x₁
POL(H₃(x₁, x₂, x₃)) = 1 + 2·x₂
POL(a) = 0
POL(f₂(x₁, x₂)) = 0
POL(g₁(x₁)) = 2 + 2·x₁
POL(h₃(x₁, x₂, x₃)) = 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:

H₃(g₁(x), y, z) -> H₃(x, y, z)

The TRS R consists of the following rules:

f₂(a, g₁(y)) -> g₁(g₁(y))
f₂(g₁(x), a) -> f₂(x, g₁(a))
f₂(g₁(x), g₁(y)) -> h₃(g₁(y), x, g₁(y))
h₃(g₁(x), y, z) -> f₂(y, h₃(x, y, z))
h₃(a, y, z) -> z

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.

H₃(g₁(x), y, z) -> H₃(x, y, z)

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

POL(H₃(x₁, x₂, x₃)) = x₁ + 3·x₁·x₂ + 3·x₁·x₂·x₃ + 3·x₁·x₃
POL(g₁(x₁)) = 1 + 2·x₁

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(a, g₁(y)) -> g₁(g₁(y))
f₂(g₁(x), a) -> f₂(x, g₁(a))
f₂(g₁(x), g₁(y)) -> h₃(g₁(y), x, g₁(y))
h₃(g₁(x), y, z) -> f₂(y, h₃(x, y, z))
h₃(a, y, z) -> z

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.