Termination w.r.t. Q proof of TRS/SK904.40.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(f₂(a, x), y), z) -> f₂(f₂(x, z), f₂(y, z))
f₂(f₂(b, x), y) -> x
f₂(c, y) -> y

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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