Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/RubioSLASHlescanne.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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.