Termination w.r.t. Q proof of TRS/Rubiotest4.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, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(a, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

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(c, c) -> F₂(a, a)
F₂(s₁(X), c) -> F₂(X, c)
F₂(a, b) -> F₂(s₁(a), c)
F₂(a, a) -> F₂(a, b)

The TRS R consists of the following rules:

f₂(a, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

F₂(c, c) -> F₂(a, a)
F₂(s₁(X), c) -> F₂(X, c)
F₂(a, b) -> F₂(s₁(a), c)
F₂(a, a) -> F₂(a, b)

The TRS R consists of the following rules:

f₂(a, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

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(c, c) -> F₂(a, a)

The remaining pairs can at least be oriented weakly.

F₂(s₁(X), c) -> F₂(X, c)
F₂(a, b) -> F₂(s₁(a), c)
F₂(a, a) -> F₂(a, b)

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

POL( F₂(x₁, x₂) ) = max{0, x₁ - 1}

POL( c ) = 2

POL( a ) = 0

POL( s₁(x₁) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

F₂(s₁(X), c) -> F₂(X, c)
F₂(a, b) -> F₂(s₁(a), c)
F₂(a, a) -> F₂(a, b)

The TRS R consists of the following rules:

f₂(a, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

F₂(s₁(X), c) -> F₂(X, c)

The TRS R consists of the following rules:

f₂(a, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

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(s₁(X), c) -> F₂(X, c)

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

POL( F₂(x₁, x₂) ) = max{0, x₁ - 1}

POL( s₁(x₁) ) = x₁ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

f₂(a, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

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.