Termination w.r.t. Q proof of TRS/Zantemaz07.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, x) -> f₂(b, f₂(c, x))
f₂(a, f₂(b, x)) -> f₂(b, f₂(a, x))
f₂(d, f₂(c, x)) -> f₂(d, f₂(a, x))
f₂(a, f₂(c, x)) -> f₂(c, f₂(a, x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

f₂(a, x) -> f₂(b, f₂(c, x))
f₂(a, f₂(b, x)) -> f₂(b, f₂(a, x))
f₂(d, f₂(c, x)) -> f₂(d, f₂(a, x))
f₂(a, f₂(c, x)) -> f₂(c, f₂(a, 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 with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

f₂(a, x) -> f₂(b, f₂(c, x))
f₂(a, f₂(b, x)) -> f₂(b, f₂(a, x))
f₂(d, f₂(c, x)) -> f₂(d, f₂(a, x))
f₂(a, f₂(c, x)) -> f₂(c, f₂(a, 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 oriented strictly and are deleted.

F₂(a, f₂(b, x)) -> F₂(a, x)

The remaining pairs can at least be oriented weakly.

F₂(a, f₂(c, x)) -> F₂(a, x)

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

POL( F₂(x₁, x₂) ) = max{0, x₂ - 3}

POL( f₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( b ) = 3

POL( c ) = 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

F₂(a, f₂(c, x)) -> F₂(a, x)

The TRS R consists of the following rules:

f₂(a, x) -> f₂(b, f₂(c, x))
f₂(a, f₂(b, x)) -> f₂(b, f₂(a, x))
f₂(d, f₂(c, x)) -> f₂(d, f₂(a, x))
f₂(a, f₂(c, x)) -> f₂(c, f₂(a, 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 oriented strictly and are deleted.

F₂(a, f₂(c, x)) -> F₂(a, x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F₂(x₁, x₂) ) = max{0, x₂ - 3}

POL( f₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( c ) = 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

f₂(a, x) -> f₂(b, f₂(c, x))
f₂(a, f₂(b, x)) -> f₂(b, f₂(a, x))
f₂(d, f₂(c, x)) -> f₂(d, f₂(a, x))
f₂(a, f₂(c, x)) -> f₂(c, f₂(a, 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 oriented strictly and are deleted.

F₂(d, f₂(c, x)) -> F₂(d, f₂(a, x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F₂(x₁, x₂) ) = max{0, x₂ - 3}

POL( f₂(x₁, x₂) ) = max{0, 2x₁ + x₂ - 2}

POL( c ) = 3

POL( a ) = 2

POL( b ) = 0

The following usable rules [14] were oriented:

f₂(a, f₂(c, x)) -> f₂(c, f₂(a, x))
f₂(a, x) -> f₂(b, f₂(c, x))
f₂(a, f₂(b, x)) -> f₂(b, 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, x) -> f₂(b, f₂(c, x))
f₂(a, f₂(b, x)) -> f₂(b, f₂(a, x))
f₂(d, f₂(c, x)) -> f₂(d, f₂(a, x))
f₂(a, f₂(c, x)) -> f₂(c, f₂(a, 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.