Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/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.

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

      ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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))

Used argument filtering: I₁(x1) = x1
div₂(x1, x2) = div₂(x1, x2)
DIV₂(x1, x2) = x1
i₁(x1) = i₁(x1)
e = e
Used ordering: Quasi Precedence: [div_2, i_1]

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ 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 contains 0 SCCs with 1 less node.