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:

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
          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


I(div(X, Y)) → DIV(Y, 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 remaining pairs can at least be oriented weakly.

DIV(X, e) → I(X)
Used ordering: Combined order from the following AFS and order.
I(x1)  =  x1
div(x1, x2)  =  div(x1, x2)
DIV(x1, x2)  =  x1
i(x1)  =  i(x1)
e  =  e

Recursive path order with status [2].
Quasi-Precedence:
[div2, i1]

Status:
i1: [1]
div2: [1,2]


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

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

DIV(X, e) → I(X)

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.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.