Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/ZantemaSLASHz26.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 strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.

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: Combined order from the following AFS and order.
A₁(x1) = x1
a₁(x1) = x1
f₂(x1, x2) = f₂(x1, x2)
F₂(x1, x2) = F₂(x1, x2)
b₁(x1) = x1

Lexicographic Path Order [19].
Precedence:

[f₂, F_2]

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.