Termination w.r.t. Q proof of TRS/Zantemaz27.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:

f₃(0, 1, x) -> f₃(g₁(x), g₁(x), x)
f₃(g₁(x), y, z) -> g₁(f₃(x, y, z))
f₃(x, g₁(y), z) -> g₁(f₃(x, y, z))
f₃(x, y, g₁(z)) -> g₁(f₃(x, y, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₃(0, 1, x) -> f₃(g₁(x), g₁(x), x)
f₃(g₁(x), y, z) -> g₁(f₃(x, y, z))
f₃(x, g₁(y), z) -> g₁(f₃(x, y, z))
f₃(x, y, g₁(z)) -> g₁(f₃(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:

F₃(g₁(x), y, z) -> F₃(x, y, z)
F₃(0, 1, x) -> F₃(g₁(x), g₁(x), x)
F₃(x, y, g₁(z)) -> F₃(x, y, z)
F₃(x, g₁(y), z) -> F₃(x, y, z)

The TRS R consists of the following rules:

f₃(0, 1, x) -> f₃(g₁(x), g₁(x), x)
f₃(g₁(x), y, z) -> g₁(f₃(x, y, z))
f₃(x, g₁(y), z) -> g₁(f₃(x, y, z))
f₃(x, y, g₁(z)) -> g₁(f₃(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:

F₃(g₁(x), y, z) -> F₃(x, y, z)
F₃(0, 1, x) -> F₃(g₁(x), g₁(x), x)
F₃(x, y, g₁(z)) -> F₃(x, y, z)
F₃(x, g₁(y), z) -> F₃(x, y, z)

The TRS R consists of the following rules:

f₃(0, 1, x) -> f₃(g₁(x), g₁(x), x)
f₃(g₁(x), y, z) -> g₁(f₃(x, y, z))
f₃(x, g₁(y), z) -> g₁(f₃(x, y, z))
f₃(x, y, g₁(z)) -> g₁(f₃(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.

F₃(x, y, g₁(z)) -> F₃(x, y, z)

The remaining pairs can at least be oriented weakly.

F₃(g₁(x), y, z) -> F₃(x, y, z)
F₃(0, 1, x) -> F₃(g₁(x), g₁(x), x)
F₃(x, g₁(y), z) -> F₃(x, y, z)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 1 ) = max{0, -1}

POL( 0 ) = max{0, -1}

POL( F₃(x₁, ..., x₃) ) = x₃ + 2

POL( g₁(x₁) ) = 2x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

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

F₃(g₁(x), y, z) -> F₃(x, y, z)
F₃(0, 1, x) -> F₃(g₁(x), g₁(x), x)
F₃(x, g₁(y), z) -> F₃(x, y, z)

The TRS R consists of the following rules:

f₃(0, 1, x) -> f₃(g₁(x), g₁(x), x)
f₃(g₁(x), y, z) -> g₁(f₃(x, y, z))
f₃(x, g₁(y), z) -> g₁(f₃(x, y, z))
f₃(x, y, g₁(z)) -> g₁(f₃(x, y, z))

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