Termination w.r.t. Q proof of TRS/Zantemaz21.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(f₂(a, b), x) -> f₂(a, f₂(a, x))
f₂(f₂(b, a), x) -> f₂(b, f₂(b, x))
f₂(x, f₂(y, z)) -> f₂(f₂(x, y), z)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

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

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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(x, f₂(y, z)) -> F₂(x, y)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

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

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

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

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

The TRS R consists of the following rules:

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

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