Termination w.r.t. Q proof of TRS/Endrullislinear2.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(a, f₂(a, f₂(x, b)))) -> f₂(f₂(a, f₂(a, f₂(a, x))), b)
f₂(f₂(f₂(a, x), b), b) -> f₂(f₂(a, f₂(f₂(x, b), b)), b)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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 with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

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

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(F₂(x₁, x₂)) = x₂
POL(a) = 3
POL(b) = 0
POL(f₂(x₁, x₂)) = x₁ + 2·x₂

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

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

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

The TRS R consists of the following rules:

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

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