Termination w.r.t. Q proof of TRS/Zantemaz20.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(b, f₂(a, f₂(c, f₂(b, f₂(a, 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(b, f₂(a, f₂(c, f₂(b, f₂(a, 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(a, b), x) -> F₂(b, f₂(a, f₂(c, f₂(b, f₂(a, x)))))
F₂(f₂(a, b), x) -> F₂(a, x)
F₂(f₂(a, b), x) -> F₂(a, f₂(c, f₂(b, f₂(a, x))))
F₂(x, f₂(y, z)) -> F₂(f₂(x, y), z)
F₂(f₂(a, b), x) -> F₂(b, f₂(a, x))
F₂(f₂(a, b), x) -> F₂(c, f₂(b, f₂(a, x)))

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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