Termination w.r.t. Q proof of TRS/various25.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:

g₁(a) -> g₁(b)
b -> f₂(a, a)
f₂(a, a) -> g₁(d)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

g₁(a) -> g₁(b)
b -> f₂(a, a)
f₂(a, a) -> g₁(d)

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, a) -> G₁(d)
B -> F₂(a, a)
G₁(a) -> B
G₁(a) -> G₁(b)

The TRS R consists of the following rules:

g₁(a) -> g₁(b)
b -> f₂(a, a)
f₂(a, a) -> g₁(d)

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, a) -> G₁(d)
B -> F₂(a, a)
G₁(a) -> B
G₁(a) -> G₁(b)

The TRS R consists of the following rules:

g₁(a) -> g₁(b)
b -> f₂(a, a)
f₂(a, a) -> g₁(d)

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 4 less nodes.