Termination w.r.t. Q proof of TRS/Zantemaz03.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:

a₂(f, a₂(g, a₂(f, x))) -> a₂(f, a₂(g, a₂(g, a₂(f, x))))
a₂(g, a₂(f, a₂(g, x))) -> a₂(g, a₂(f, a₂(f, a₂(g, x))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a₂(f, a₂(g, a₂(f, x))) -> a₂(f, a₂(g, a₂(g, a₂(f, x))))
a₂(g, a₂(f, a₂(g, x))) -> a₂(g, a₂(f, a₂(f, a₂(g, x))))

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:

A₂(g, a₂(f, a₂(g, x))) -> A₂(g, a₂(f, a₂(f, a₂(g, x))))
A₂(f, a₂(g, a₂(f, x))) -> A₂(g, a₂(g, a₂(f, x)))
A₂(g, a₂(f, a₂(g, x))) -> A₂(f, a₂(f, a₂(g, x)))
A₂(f, a₂(g, a₂(f, x))) -> A₂(f, a₂(g, a₂(g, a₂(f, x))))

The TRS R consists of the following rules:

a₂(f, a₂(g, a₂(f, x))) -> a₂(f, a₂(g, a₂(g, a₂(f, x))))
a₂(g, a₂(f, a₂(g, x))) -> a₂(g, a₂(f, a₂(f, a₂(g, x))))

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:

A₂(g, a₂(f, a₂(g, x))) -> A₂(g, a₂(f, a₂(f, a₂(g, x))))
A₂(f, a₂(g, a₂(f, x))) -> A₂(g, a₂(g, a₂(f, x)))
A₂(g, a₂(f, a₂(g, x))) -> A₂(f, a₂(f, a₂(g, x)))
A₂(f, a₂(g, a₂(f, x))) -> A₂(f, a₂(g, a₂(g, a₂(f, x))))

The TRS R consists of the following rules:

a₂(f, a₂(g, a₂(f, x))) -> a₂(f, a₂(g, a₂(g, a₂(f, x))))
a₂(g, a₂(f, a₂(g, x))) -> a₂(g, a₂(f, a₂(f, a₂(g, x))))

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