Termination w.r.t. Q proof of TRS/various26.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(f₁(x)) -> f₁(g₂(f₁(x), x))
f₁(f₁(x)) -> f₁(h₂(f₁(x), f₁(x)))
g₂(x, y) -> y
h₂(x, x) -> g₂(x, 0)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(f₁(x)) -> f₁(g₂(f₁(x), x))
f₁(f₁(x)) -> f₁(h₂(f₁(x), f₁(x)))
g₂(x, y) -> y
h₂(x, x) -> g₂(x, 0)

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(f₁(x)) -> F₁(g₂(f₁(x), x))
F₁(f₁(x)) -> H₂(f₁(x), f₁(x))
H₂(x, x) -> G₂(x, 0)
F₁(f₁(x)) -> G₂(f₁(x), x)
F₁(f₁(x)) -> F₁(h₂(f₁(x), f₁(x)))

The TRS R consists of the following rules:

f₁(f₁(x)) -> f₁(g₂(f₁(x), x))
f₁(f₁(x)) -> f₁(h₂(f₁(x), f₁(x)))
g₂(x, y) -> y
h₂(x, x) -> g₂(x, 0)

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(f₁(x)) -> F₁(g₂(f₁(x), x))
F₁(f₁(x)) -> H₂(f₁(x), f₁(x))
H₂(x, x) -> G₂(x, 0)
F₁(f₁(x)) -> G₂(f₁(x), x)
F₁(f₁(x)) -> F₁(h₂(f₁(x), f₁(x)))

The TRS R consists of the following rules:

f₁(f₁(x)) -> f₁(g₂(f₁(x), x))
f₁(f₁(x)) -> f₁(h₂(f₁(x), f₁(x)))
g₂(x, y) -> y
h₂(x, x) -> g₂(x, 0)

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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

F₁(f₁(x)) -> F₁(g₂(f₁(x), x))
F₁(f₁(x)) -> F₁(h₂(f₁(x), f₁(x)))

The TRS R consists of the following rules:

f₁(f₁(x)) -> f₁(g₂(f₁(x), x))
f₁(f₁(x)) -> f₁(h₂(f₁(x), f₁(x)))
g₂(x, y) -> y
h₂(x, x) -> g₂(x, 0)

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