Termination w.r.t. Q proof of TRS/nontermin/caribooex4.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(h₁(x)) -> f₁(i₁(x))
f₁(i₁(x)) -> a
i₁(x) -> h₁(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(h₁(x)) -> F₁(i₁(x))
F₁(h₁(x)) -> I₁(x)

The TRS R consists of the following rules:

f₁(h₁(x)) -> f₁(i₁(x))
f₁(i₁(x)) -> a
i₁(x) -> h₁(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(h₁(x)) -> F₁(i₁(x))
F₁(h₁(x)) -> I₁(x)

The TRS R consists of the following rules:

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

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

F₁(h₁(x)) -> F₁(i₁(x))

The TRS R consists of the following rules:

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

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