Termination w.r.t. Q proof of TRS/Rubiop266.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:

f₁(f₁(X)) -> f₁(a₁(b₁(f₁(X))))
f₁(a₁(g₁(X))) -> b₁(X)
b₁(X) -> a₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(f₁(X)) -> f₁(a₁(b₁(f₁(X))))
f₁(a₁(g₁(X))) -> b₁(X)
b₁(X) -> a₁(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:

F₁(f₁(X)) -> B₁(f₁(X))
F₁(a₁(g₁(X))) -> B₁(X)
F₁(f₁(X)) -> F₁(a₁(b₁(f₁(X))))

The TRS R consists of the following rules:

f₁(f₁(X)) -> f₁(a₁(b₁(f₁(X))))
f₁(a₁(g₁(X))) -> b₁(X)
b₁(X) -> a₁(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:

F₁(f₁(X)) -> B₁(f₁(X))
F₁(a₁(g₁(X))) -> B₁(X)
F₁(f₁(X)) -> F₁(a₁(b₁(f₁(X))))

The TRS R consists of the following rules:

f₁(f₁(X)) -> f₁(a₁(b₁(f₁(X))))
f₁(a₁(g₁(X))) -> b₁(X)
b₁(X) -> a₁(X)

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.