Termination w.r.t. Q proof of TRS/D3327.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

h₂(f₁(x), y) -> f₁(g₂(x, y))
g₂(x, y) -> h₂(x, 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(f₁(x), y) -> G₂(x, y)

The remaining pairs can at least be oriented weakly.

G₂(x, y) -> H₂(x, y)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( G₂(x₁, x₂) ) = 2x₁

POL( H₂(x₁, x₂) ) = 2x₁

POL( f₁(x₁) ) = x₁ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

G₂(x, y) -> H₂(x, y)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.