Termination w.r.t. Q proof of TRS/Zantemaz06.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, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, 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:

F₂(b, f₂(a, x)) -> F₂(b, f₂(b, f₂(b, x)))
F₂(a, f₂(b, x)) -> F₂(a, x)
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, x))
F₂(b, f₂(a, x)) -> F₂(b, x)
F₂(b, f₂(a, x)) -> F₂(b, f₂(b, x))
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, f₂(a, x)))

The TRS R consists of the following rules:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

F₂(b, f₂(a, x)) -> F₂(b, f₂(b, f₂(b, x)))
F₂(a, f₂(b, x)) -> F₂(a, x)
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, x))
F₂(b, f₂(a, x)) -> F₂(b, x)
F₂(b, f₂(a, x)) -> F₂(b, f₂(b, x))
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, f₂(a, x)))

The TRS R consists of the following rules:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

F₂(b, f₂(a, x)) -> F₂(b, f₂(b, f₂(b, x)))
F₂(b, f₂(a, x)) -> F₂(b, x)
F₂(b, f₂(a, x)) -> F₂(b, f₂(b, x))

The TRS R consists of the following rules:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, 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.

F₂(b, f₂(a, x)) -> F₂(b, x)
F₂(b, f₂(a, x)) -> F₂(b, f₂(b, x))

The remaining pairs can at least be oriented weakly.

F₂(b, f₂(a, x)) -> F₂(b, f₂(b, f₂(b, x)))

Used ordering: Polynomial interpretation [21]:

POL(F₂(x₁, x₂)) = x₂
POL(a) = 2
POL(b) = 1
POL(f₂(x₁, x₂)) = x₁ + x₂

The following usable rules [14] were oriented:

f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

F₂(b, f₂(a, x)) -> F₂(b, f₂(b, f₂(b, x)))

The TRS R consists of the following rules:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, 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.

F₂(b, f₂(a, x)) -> F₂(b, f₂(b, f₂(b, x)))

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

POL(F₂(x₁, x₂)) = 2·x₂
POL(a) = 2
POL(b) = 0
POL(f₂(x₁, x₂)) = 2 + 2·x₁

The following usable rules [14] were oriented:

f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, 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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

F₂(a, f₂(b, x)) -> F₂(a, x)
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, x))
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, f₂(a, x)))

The TRS R consists of the following rules:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, 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.

F₂(a, f₂(b, x)) -> F₂(a, x)
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, x))
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, f₂(a, x)))

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

POL(F₂(x₁, x₂)) = x₂
POL(a) = 0
POL(b) = 1
POL(f₂(x₁, x₂)) = x₁ + x₂

The following usable rules [14] were oriented:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, 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.