Termination w.r.t. Q proof of TRS/Zantemaz25.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(F₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(a) = 0
POL(b) = 0
POL(c) = 1
POL(f₂(x₁, x₂)) = x₁ + x₂

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(F₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(a) = 1
POL(b) = 0
POL(c) = 1
POL(f₂(x₁, x₂)) = x₁ + x₂

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

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

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

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

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

The TRS R consists of the following rules:

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

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