Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/ZantemaSLASHz06.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, 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:

F₂(b, f₂(a, x)) -> F₂(b, f₂(b, f₂(b, x)))
F₂(a, f₂(b, x)) -> F₂(a, x)
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, x))
F₂(b, f₂(a, x)) -> F₂(b, x)
F₂(b, f₂(a, x)) -> F₂(b, f₂(b, x))
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, f₂(a, x)))

The TRS R consists of the following rules:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, 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:

F₂(b, f₂(a, x)) -> F₂(b, f₂(b, f₂(b, x)))
F₂(a, f₂(b, x)) -> F₂(a, x)
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, x))
F₂(b, f₂(a, x)) -> F₂(b, x)
F₂(b, f₂(a, x)) -> F₂(b, f₂(b, x))
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, f₂(a, x)))

The TRS R consists of the following rules:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

F₂(b, f₂(a, x)) -> F₂(b, f₂(b, f₂(b, x)))
F₂(b, f₂(a, x)) -> F₂(b, x)
F₂(b, f₂(a, x)) -> F₂(b, f₂(b, x))

The TRS R consists of the following rules:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, x)))

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.

F₂(b, f₂(a, x)) -> F₂(b, f₂(b, f₂(b, x)))
F₂(b, f₂(a, x)) -> F₂(b, x)
F₂(b, f₂(a, x)) -> F₂(b, f₂(b, x))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F₂(x1, x2) = F₁(x2)
b = b
f₂(x1, x2) = f₂(x1, x2)
a = a

Lexicographic Path Order [19].
Precedence:

f₂ > [F₁, b]
a > [F₁, b]

The following usable rules [14] were oriented:

f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

F₂(a, f₂(b, x)) -> F₂(a, x)
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, x))
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, f₂(a, x)))

The TRS R consists of the following rules:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, x)))

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.

F₂(a, f₂(b, x)) -> F₂(a, x)
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, x))
F₂(a, f₂(b, x)) -> F₂(a, f₂(a, f₂(a, x)))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F₂(x1, x2) = F₁(x2)
a = a
f₂(x1, x2) = f₂(x1, x2)
b = b

Lexicographic Path Order [19].
Precedence:

F₁ > [a, f_2]
b > [a, f_2]

The following usable rules [14] were oriented:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f₂(a, f₂(b, x)) -> f₂(a, f₂(a, f₂(a, x)))
f₂(b, f₂(a, x)) -> f₂(b, f₂(b, f₂(b, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.