Termination w.r.t. Q proof of TRS/SK902.03.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:

minus₁(minus₁(x)) -> x
minus₁(h₁(x)) -> h₁(minus₁(x))
minus₁(f₂(x, y)) -> f₂(minus₁(y), minus₁(x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus₁(minus₁(x)) -> x
minus₁(h₁(x)) -> h₁(minus₁(x))
minus₁(f₂(x, y)) -> f₂(minus₁(y), minus₁(x))

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:

MINUS₁(h₁(x)) -> MINUS₁(x)
MINUS₁(f₂(x, y)) -> MINUS₁(x)
MINUS₁(f₂(x, y)) -> MINUS₁(y)

The TRS R consists of the following rules:

minus₁(minus₁(x)) -> x
minus₁(h₁(x)) -> h₁(minus₁(x))
minus₁(f₂(x, y)) -> f₂(minus₁(y), minus₁(x))

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:

MINUS₁(h₁(x)) -> MINUS₁(x)
MINUS₁(f₂(x, y)) -> MINUS₁(x)
MINUS₁(f₂(x, y)) -> MINUS₁(y)

The TRS R consists of the following rules:

minus₁(minus₁(x)) -> x
minus₁(h₁(x)) -> h₁(minus₁(x))
minus₁(f₂(x, y)) -> f₂(minus₁(y), minus₁(x))

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.

MINUS₁(h₁(x)) -> MINUS₁(x)

The remaining pairs can at least be oriented weakly.

MINUS₁(f₂(x, y)) -> MINUS₁(x)
MINUS₁(f₂(x, y)) -> MINUS₁(y)

Used ordering: Polynomial interpretation [21]:

POL(MINUS₁(x₁)) = x₁
POL(f₂(x₁, x₂)) = 3·x₁ + 3·x₂
POL(h₁(x₁)) = 1 + 2·x₁

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:

MINUS₁(f₂(x, y)) -> MINUS₁(x)
MINUS₁(f₂(x, y)) -> MINUS₁(y)

The TRS R consists of the following rules:

minus₁(minus₁(x)) -> x
minus₁(h₁(x)) -> h₁(minus₁(x))
minus₁(f₂(x, y)) -> f₂(minus₁(y), minus₁(x))

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.

MINUS₁(f₂(x, y)) -> MINUS₁(x)
MINUS₁(f₂(x, y)) -> MINUS₁(y)

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

POL(MINUS₁(x₁)) = 2·x₁
POL(f₂(x₁, x₂)) = 1 + 2·x₁ + x₂

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:

minus₁(minus₁(x)) -> x
minus₁(h₁(x)) -> h₁(minus₁(x))
minus₁(f₂(x, y)) -> f₂(minus₁(y), minus₁(x))

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.