Termination w.r.t. Q proof of TRS/HMt011.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:

g₂(f₁(x), y) -> f₁(h₂(x, y))
h₂(x, y) -> g₂(x, f₁(y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

g₂(f₁(x), y) -> f₁(h₂(x, y))
h₂(x, y) -> g₂(x, f₁(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:

H₂(x, y) -> G₂(x, f₁(y))
G₂(f₁(x), y) -> H₂(x, y)

The TRS R consists of the following rules:

g₂(f₁(x), y) -> f₁(h₂(x, y))
h₂(x, y) -> g₂(x, f₁(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:

H₂(x, y) -> G₂(x, f₁(y))
G₂(f₁(x), y) -> H₂(x, y)

The TRS R consists of the following rules:

g₂(f₁(x), y) -> f₁(h₂(x, y))
h₂(x, y) -> g₂(x, f₁(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.

H₂(x, y) -> G₂(x, f₁(y))
G₂(f₁(x), y) -> H₂(x, y)

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

POL(G₂(x₁, x₂)) = 1 + 2·x₁
POL(H₂(x₁, x₂)) = 2 + 2·x₁
POL(f₁(x₁)) = 2 + 2·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:

g₂(f₁(x), y) -> f₁(h₂(x, y))
h₂(x, y) -> g₂(x, f₁(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.