Termination w.r.t. Q proof of TRS/HMt007.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(a) -> f₁(b)
g₁(b) -> g₁(a)
f₁(x) -> g₁(x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(a) -> f₁(b)
g₁(b) -> g₁(a)
f₁(x) -> g₁(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:

G₁(b) -> G₁(a)
F₁(x) -> G₁(x)
F₁(a) -> F₁(b)

The TRS R consists of the following rules:

f₁(a) -> f₁(b)
g₁(b) -> g₁(a)
f₁(x) -> g₁(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:

G₁(b) -> G₁(a)
F₁(x) -> G₁(x)
F₁(a) -> F₁(b)

The TRS R consists of the following rules:

f₁(a) -> f₁(b)
g₁(b) -> g₁(a)
f₁(x) -> g₁(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.