Termination w.r.t. Q proof of TRS/Zantemaz24.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(x, a₁(b₁(y))) -> f₂(a₁(b₁(x)), y)
f₂(x, b₁(c₁(y))) -> f₂(b₁(c₁(x)), y)
f₂(x, c₁(a₁(y))) -> f₂(c₁(a₁(x)), y)
f₂(a₁(x), y) -> f₂(x, a₁(y))
f₂(b₁(x), y) -> f₂(x, b₁(y))
f₂(c₁(x), y) -> f₂(x, c₁(y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(x, a₁(b₁(y))) -> f₂(a₁(b₁(x)), y)
f₂(x, b₁(c₁(y))) -> f₂(b₁(c₁(x)), y)
f₂(x, c₁(a₁(y))) -> f₂(c₁(a₁(x)), y)
f₂(a₁(x), y) -> f₂(x, a₁(y))
f₂(b₁(x), y) -> f₂(x, b₁(y))
f₂(c₁(x), y) -> f₂(x, c₁(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(x, c₁(a₁(y))) -> F₂(c₁(a₁(x)), y)
F₂(b₁(x), y) -> F₂(x, b₁(y))
F₂(a₁(x), y) -> F₂(x, a₁(y))
F₂(x, a₁(b₁(y))) -> F₂(a₁(b₁(x)), y)
F₂(x, b₁(c₁(y))) -> F₂(b₁(c₁(x)), y)
F₂(c₁(x), y) -> F₂(x, c₁(y))

The TRS R consists of the following rules:

f₂(x, a₁(b₁(y))) -> f₂(a₁(b₁(x)), y)
f₂(x, b₁(c₁(y))) -> f₂(b₁(c₁(x)), y)
f₂(x, c₁(a₁(y))) -> f₂(c₁(a₁(x)), y)
f₂(a₁(x), y) -> f₂(x, a₁(y))
f₂(b₁(x), y) -> f₂(x, b₁(y))
f₂(c₁(x), y) -> f₂(x, c₁(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(x, c₁(a₁(y))) -> F₂(c₁(a₁(x)), y)
F₂(b₁(x), y) -> F₂(x, b₁(y))
F₂(a₁(x), y) -> F₂(x, a₁(y))
F₂(x, a₁(b₁(y))) -> F₂(a₁(b₁(x)), y)
F₂(x, b₁(c₁(y))) -> F₂(b₁(c₁(x)), y)
F₂(c₁(x), y) -> F₂(x, c₁(y))

The TRS R consists of the following rules:

f₂(x, a₁(b₁(y))) -> f₂(a₁(b₁(x)), y)
f₂(x, b₁(c₁(y))) -> f₂(b₁(c₁(x)), y)
f₂(x, c₁(a₁(y))) -> f₂(c₁(a₁(x)), y)
f₂(a₁(x), y) -> f₂(x, a₁(y))
f₂(b₁(x), y) -> f₂(x, b₁(y))
f₂(c₁(x), y) -> f₂(x, c₁(y))

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