Termination w.r.t. Q proof of TRS/SK904.44.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(h₁(x)) -> f₁(i₁(x))
g₁(i₁(x)) -> g₁(h₁(x))
h₁(a) -> b
i₁(a) -> b

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(h₁(x)) -> f₁(i₁(x))
g₁(i₁(x)) -> g₁(h₁(x))
h₁(a) -> b
i₁(a) -> b

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(i₁(x)) -> H₁(x)
F₁(h₁(x)) -> F₁(i₁(x))
F₁(h₁(x)) -> I₁(x)
G₁(i₁(x)) -> G₁(h₁(x))

The TRS R consists of the following rules:

f₁(h₁(x)) -> f₁(i₁(x))
g₁(i₁(x)) -> g₁(h₁(x))
h₁(a) -> b
i₁(a) -> b

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(i₁(x)) -> H₁(x)
F₁(h₁(x)) -> F₁(i₁(x))
F₁(h₁(x)) -> I₁(x)
G₁(i₁(x)) -> G₁(h₁(x))

The TRS R consists of the following rules:

f₁(h₁(x)) -> f₁(i₁(x))
g₁(i₁(x)) -> g₁(h₁(x))
h₁(a) -> b
i₁(a) -> b

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.