Termination w.r.t. Q proof of TRS/nontermin/caribooex6.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f₂(x, x) -> f₂(i₁(x), g₁(g₁(x)))
f₂(x, y) -> x
g₁(x) -> i₁(x)
f₂(x, i₁(x)) -> f₂(x, x)
f₂(i₁(x), i₁(g₁(x))) -> a

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(x, x) -> f₂(i₁(x), g₁(g₁(x)))
f₂(x, y) -> x
g₁(x) -> i₁(x)
f₂(x, i₁(x)) -> f₂(x, x)
f₂(i₁(x), i₁(g₁(x))) -> a

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(x, x) -> F₂(i₁(x), g₁(g₁(x)))
F₂(x, x) -> G₁(x)
F₂(x, i₁(x)) -> F₂(x, x)
F₂(x, x) -> G₁(g₁(x))

The TRS R consists of the following rules:

f₂(x, x) -> f₂(i₁(x), g₁(g₁(x)))
f₂(x, y) -> x
g₁(x) -> i₁(x)
f₂(x, i₁(x)) -> f₂(x, x)
f₂(i₁(x), i₁(g₁(x))) -> a

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(x, x) -> F₂(i₁(x), g₁(g₁(x)))
F₂(x, x) -> G₁(x)
F₂(x, i₁(x)) -> F₂(x, x)
F₂(x, x) -> G₁(g₁(x))

The TRS R consists of the following rules:

f₂(x, x) -> f₂(i₁(x), g₁(g₁(x)))
f₂(x, y) -> x
g₁(x) -> i₁(x)
f₂(x, i₁(x)) -> f₂(x, x)
f₂(i₁(x), i₁(g₁(x))) -> a

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

F₂(x, x) -> F₂(i₁(x), g₁(g₁(x)))
F₂(x, i₁(x)) -> F₂(x, x)

The TRS R consists of the following rules:

f₂(x, x) -> f₂(i₁(x), g₁(g₁(x)))
f₂(x, y) -> x
g₁(x) -> i₁(x)
f₂(x, i₁(x)) -> f₂(x, x)
f₂(i₁(x), i₁(g₁(x))) -> a

Q is empty.
We have to consider all minimal (P,Q,R)-chains.