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 Order [17,21] with Interpretation:

POL( H₃(x₁, ..., x₃) ) = 3x₂ + 3

POL( F₂(x₁, x₂) ) = max{0, 3x₁ - 1}

POL( h₃(x₁, ..., x₃) ) = max{0, 3x₁ + 3x₂ - 1}

POL( a ) = 1

POL( g₁(x₁) ) = x₁ + 2

POL( f₂(x₁, x₂) ) = max{0, -3}

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 Order [17,21] with Interpretation:

POL( H₃(x₁, ..., x₃) ) = 2x₁ + 2x₂ + 3x₃ + 2

POL( g₁(x₁) ) = 2x₁ + 1

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.