Termination w.r.t. Q proof of TRS/Zantemaz26.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:

a₁(a₁(f₂(x, y))) -> f₂(a₁(b₁(a₁(b₁(a₁(x))))), a₁(b₁(a₁(b₁(a₁(y))))))
f₂(a₁(x), a₁(y)) -> a₁(f₂(x, y))
f₂(b₁(x), b₁(y)) -> b₁(f₂(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a₁(a₁(f₂(x, y))) -> f₂(a₁(b₁(a₁(b₁(a₁(x))))), a₁(b₁(a₁(b₁(a₁(y))))))
f₂(a₁(x), a₁(y)) -> a₁(f₂(x, y))
f₂(b₁(x), b₁(y)) -> b₁(f₂(x, y))

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:

A₁(a₁(f₂(x, y))) -> A₁(y)
F₂(a₁(x), a₁(y)) -> F₂(x, y)
A₁(a₁(f₂(x, y))) -> A₁(b₁(a₁(y)))
A₁(a₁(f₂(x, y))) -> A₁(b₁(a₁(b₁(a₁(x)))))
A₁(a₁(f₂(x, y))) -> A₁(x)
F₂(b₁(x), b₁(y)) -> F₂(x, y)
F₂(a₁(x), a₁(y)) -> A₁(f₂(x, y))
A₁(a₁(f₂(x, y))) -> A₁(b₁(a₁(b₁(a₁(y)))))
A₁(a₁(f₂(x, y))) -> F₂(a₁(b₁(a₁(b₁(a₁(x))))), a₁(b₁(a₁(b₁(a₁(y))))))
A₁(a₁(f₂(x, y))) -> A₁(b₁(a₁(x)))

The TRS R consists of the following rules:

a₁(a₁(f₂(x, y))) -> f₂(a₁(b₁(a₁(b₁(a₁(x))))), a₁(b₁(a₁(b₁(a₁(y))))))
f₂(a₁(x), a₁(y)) -> a₁(f₂(x, y))
f₂(b₁(x), b₁(y)) -> b₁(f₂(x, y))

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:

A₁(a₁(f₂(x, y))) -> A₁(y)
F₂(a₁(x), a₁(y)) -> F₂(x, y)
A₁(a₁(f₂(x, y))) -> A₁(b₁(a₁(y)))
A₁(a₁(f₂(x, y))) -> A₁(b₁(a₁(b₁(a₁(x)))))
A₁(a₁(f₂(x, y))) -> A₁(x)
F₂(b₁(x), b₁(y)) -> F₂(x, y)
F₂(a₁(x), a₁(y)) -> A₁(f₂(x, y))
A₁(a₁(f₂(x, y))) -> A₁(b₁(a₁(b₁(a₁(y)))))
A₁(a₁(f₂(x, y))) -> F₂(a₁(b₁(a₁(b₁(a₁(x))))), a₁(b₁(a₁(b₁(a₁(y))))))
A₁(a₁(f₂(x, y))) -> A₁(b₁(a₁(x)))

The TRS R consists of the following rules:

a₁(a₁(f₂(x, y))) -> f₂(a₁(b₁(a₁(b₁(a₁(x))))), a₁(b₁(a₁(b₁(a₁(y))))))
f₂(a₁(x), a₁(y)) -> a₁(f₂(x, y))
f₂(b₁(x), b₁(y)) -> b₁(f₂(x, y))

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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

A₁(a₁(f₂(x, y))) -> A₁(y)
F₂(a₁(x), a₁(y)) -> F₂(x, y)
A₁(a₁(f₂(x, y))) -> A₁(x)
F₂(b₁(x), b₁(y)) -> F₂(x, y)
F₂(a₁(x), a₁(y)) -> A₁(f₂(x, y))
A₁(a₁(f₂(x, y))) -> F₂(a₁(b₁(a₁(b₁(a₁(x))))), a₁(b₁(a₁(b₁(a₁(y))))))

The TRS R consists of the following rules:

a₁(a₁(f₂(x, y))) -> f₂(a₁(b₁(a₁(b₁(a₁(x))))), a₁(b₁(a₁(b₁(a₁(y))))))
f₂(a₁(x), a₁(y)) -> a₁(f₂(x, y))
f₂(b₁(x), b₁(y)) -> b₁(f₂(x, y))

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