Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/RubioSLASHlescanne.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 strictly oriented and are deleted.

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 remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
I₁(x1) = I₁(x1)
div₂(x1, x2) = div₂(x1, x2)
DIV₂(x1, x2) = DIV₁(x1)
e = e
i₁(x1) = i₁(x1)

Lexicographic Path Order [19].
Precedence:

[div₂, e, i_1] > DIV₁ > I₁

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

          ↳ PisEmptyProof

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.