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