Termination w.r.t. Q proof of TRS/nontermin/caribootricky1.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:

f₂(g₁(x), g₁(y)) -> f₂(p₁(f₂(g₁(x), s₁(y))), g₁(s₁(p₁(x))))
p₁(0) -> g₁(0)
g₁(s₁(p₁(x))) -> p₁(x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(g₁(x), g₁(y)) -> f₂(p₁(f₂(g₁(x), s₁(y))), g₁(s₁(p₁(x))))
p₁(0) -> g₁(0)
g₁(s₁(p₁(x))) -> p₁(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:

F₂(g₁(x), g₁(y)) -> P₁(f₂(g₁(x), s₁(y)))
F₂(g₁(x), g₁(y)) -> F₂(p₁(f₂(g₁(x), s₁(y))), g₁(s₁(p₁(x))))
F₂(g₁(x), g₁(y)) -> F₂(g₁(x), s₁(y))
P₁(0) -> G₁(0)
F₂(g₁(x), g₁(y)) -> G₁(s₁(p₁(x)))
F₂(g₁(x), g₁(y)) -> P₁(x)

The TRS R consists of the following rules:

f₂(g₁(x), g₁(y)) -> f₂(p₁(f₂(g₁(x), s₁(y))), g₁(s₁(p₁(x))))
p₁(0) -> g₁(0)
g₁(s₁(p₁(x))) -> p₁(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:

F₂(g₁(x), g₁(y)) -> P₁(f₂(g₁(x), s₁(y)))
F₂(g₁(x), g₁(y)) -> F₂(p₁(f₂(g₁(x), s₁(y))), g₁(s₁(p₁(x))))
F₂(g₁(x), g₁(y)) -> F₂(g₁(x), s₁(y))
P₁(0) -> G₁(0)
F₂(g₁(x), g₁(y)) -> G₁(s₁(p₁(x)))
F₂(g₁(x), g₁(y)) -> P₁(x)

The TRS R consists of the following rules:

f₂(g₁(x), g₁(y)) -> f₂(p₁(f₂(g₁(x), s₁(y))), g₁(s₁(p₁(x))))
p₁(0) -> g₁(0)
g₁(s₁(p₁(x))) -> p₁(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 6 less nodes.