Termination w.r.t. Q proof of TRS/Rubiolescanne.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:

div₂(X, e) -> i₁(X)
i₁(div₂(X, Y)) -> div₂(Y, X)
div₂(div₂(X, Y), Z) -> div₂(Y, div₂(i₁(X), Z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

div₂(X, e) -> i₁(X)
i₁(div₂(X, Y)) -> div₂(Y, X)
div₂(div₂(X, Y), Z) -> div₂(Y, div₂(i₁(X), Z))

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:

I₁(div₂(X, Y)) -> DIV₂(Y, X)
DIV₂(X, e) -> I₁(X)
DIV₂(div₂(X, Y), Z) -> I₁(X)
DIV₂(div₂(X, Y), Z) -> DIV₂(i₁(X), Z)
DIV₂(div₂(X, Y), Z) -> DIV₂(Y, div₂(i₁(X), Z))

The TRS R consists of the following rules:

div₂(X, e) -> i₁(X)
i₁(div₂(X, Y)) -> div₂(Y, X)
div₂(div₂(X, Y), Z) -> div₂(Y, div₂(i₁(X), Z))

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:

I₁(div₂(X, Y)) -> DIV₂(Y, X)
DIV₂(X, e) -> I₁(X)
DIV₂(div₂(X, Y), Z) -> I₁(X)
DIV₂(div₂(X, Y), Z) -> DIV₂(i₁(X), Z)
DIV₂(div₂(X, Y), Z) -> DIV₂(Y, div₂(i₁(X), Z))

The TRS R consists of the following rules:

div₂(X, e) -> i₁(X)
i₁(div₂(X, Y)) -> div₂(Y, X)
div₂(div₂(X, Y), Z) -> div₂(Y, div₂(i₁(X), Z))

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.

I₁(div₂(X, Y)) -> DIV₂(Y, X)
DIV₂(div₂(X, Y), Z) -> I₁(X)
DIV₂(div₂(X, Y), Z) -> DIV₂(i₁(X), Z)
DIV₂(div₂(X, Y), Z) -> DIV₂(Y, div₂(i₁(X), Z))

The remaining pairs can at least be oriented weakly.

DIV₂(X, e) -> I₁(X)

Used ordering: Polynomial interpretation [21]:

POL(DIV₂(x₁, x₂)) = 3·x₁ + 2·x₁·x₂ + 2·x₂
POL(I₁(x₁)) = 2·x₁
POL(div₂(x₁, x₂)) = 3 + 3·x₁ + 2·x₁·x₂ + 3·x₂
POL(e) = 0
POL(i₁(x₁)) = x₁

The following usable rules [14] were oriented:

div₂(div₂(X, Y), Z) -> div₂(Y, div₂(i₁(X), Z))
div₂(X, e) -> i₁(X)
i₁(div₂(X, Y)) -> div₂(Y, X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

DIV₂(X, e) -> I₁(X)

The TRS R consists of the following rules:

div₂(X, e) -> i₁(X)
i₁(div₂(X, Y)) -> div₂(Y, X)
div₂(div₂(X, Y), Z) -> div₂(Y, div₂(i₁(X), Z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.