Termination w.r.t. Q proof of TRS/Zantemaz30.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(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(a, f₂(b, x)))))))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(b, f₂(a, f₂(a, f₂(a, f₂(b, x))))))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(b, x)
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(b, f₂(a, f₂(a, f₂(a, f₂(b, x)))))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(b, x))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(a, f₂(a, f₂(b, x))))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(b, f₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(a, f₂(b, x))))))))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(a, f₂(b, x)))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(a, f₂(b, x)))))))))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(a, f₂(b, x)))))))

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(b, f₂(a, f₂(a, f₂(a, f₂(b, x))))))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(b, x)
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(b, f₂(a, f₂(a, f₂(a, f₂(b, x)))))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(b, x))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(a, f₂(a, f₂(b, x))))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(b, f₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(a, f₂(b, x))))))))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(a, f₂(b, x)))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(a, f₂(b, x)))))))))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(a, f₂(b, x)))))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(a, f₂(a, f₂(b, x))))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(a, f₂(b, x)))
F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(b, f₂(a, x))))))) -> F₂(a, f₂(a, f₂(b, f₂(a, f₂(a, f₂(a, f₂(b, x)))))))

The TRS R consists of the following rules:

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

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