Termination w.r.t. Q proof of TRS/nontermin/AG01#4.21.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(1) -> f₁(g₁(1))
f₁(f₁(x)) -> f₁(x)
g₁(0) -> g₁(f₁(0))
g₁(g₁(x)) -> g₁(x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(1) -> f₁(g₁(1))
f₁(f₁(x)) -> f₁(x)
g₁(0) -> g₁(f₁(0))
g₁(g₁(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:

F₁(1) -> F₁(g₁(1))
G₁(0) -> G₁(f₁(0))
F₁(1) -> G₁(1)
G₁(0) -> F₁(0)

The TRS R consists of the following rules:

f₁(1) -> f₁(g₁(1))
f₁(f₁(x)) -> f₁(x)
g₁(0) -> g₁(f₁(0))
g₁(g₁(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:

F₁(1) -> F₁(g₁(1))
G₁(0) -> G₁(f₁(0))
F₁(1) -> G₁(1)
G₁(0) -> F₁(0)

The TRS R consists of the following rules:

f₁(1) -> f₁(g₁(1))
f₁(f₁(x)) -> f₁(x)
g₁(0) -> g₁(f₁(0))
g₁(g₁(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 4 less nodes.