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

          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

F₂(a₁(x), a₁(y)) -> F₂(x, y)
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))))))

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F₂(x₁, x₂) ) = 2x₁ + x₂ + 1

POL( a₁(x₁) ) = x₁

POL( b₁(x₁) ) = x₁

POL( f₂(x₁, x₂) ) = 2x₁ + x₂ + 2

POL( A₁(x₁) ) = max{0, x₁ - 1}

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

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

F₂(a₁(x), a₁(y)) -> F₂(x, y)
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.