Termination w.r.t. Q proof of TRS/Zantemajw35.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 3 less nodes.