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

div(X, e) → i(X)
i(div(X, Y)) → div(Y, X)
div(div(X, Y), Z) → div(Y, div(i(X), Z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

div(X, e) → i(X)
i(div(X, Y)) → div(Y, X)
div(div(X, Y), Z) → div(Y, div(i(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:

I(div(X, Y)) → DIV(Y, X)
DIV(X, e) → I(X)
DIV(div(X, Y), Z) → I(X)
DIV(div(X, Y), Z) → DIV(i(X), Z)
DIV(div(X, Y), Z) → DIV(Y, div(i(X), Z))

The TRS R consists of the following rules:

div(X, e) → i(X)
i(div(X, Y)) → div(Y, X)
div(div(X, Y), Z) → div(Y, div(i(X), Z))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

I(div(X, Y)) → DIV(Y, X)
DIV(X, e) → I(X)
DIV(div(X, Y), Z) → I(X)
DIV(div(X, Y), Z) → DIV(i(X), Z)
DIV(div(X, Y), Z) → DIV(Y, div(i(X), Z))

The TRS R consists of the following rules:

div(X, e) → i(X)
i(div(X, Y)) → div(Y, X)
div(div(X, Y), Z) → div(Y, div(i(X), Z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

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

I(div(X, Y)) → DIV(Y, X)
DIV(X, e) → I(X)
DIV(div(X, Y), Z) → I(X)
DIV(div(X, Y), Z) → DIV(Y, div(i(X), Z))
DIV(div(X, Y), Z) → DIV(i(X), Z)

The TRS R consists of the following rules:

div(X, e) → i(X)
i(div(X, Y)) → div(Y, X)
div(div(X, Y), Z) → div(Y, div(i(X), Z))

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