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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

DIV(div(X, Y), Z) → DIV(Y, div(i(X), Z))
I(div(X, Y)) → DIV(Y, X)
DIV(div(X, Y), Z) → DIV(i(X), Z)
DIV(X, e) → I(X)
DIV(div(X, Y), Z) → 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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


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

DIV(div(X, Y), Z) → DIV(Y, div(i(X), Z))
Used ordering: Polynomial interpretation [25,35]:

POL(i(x1)) = x_1   
POL(div(x1, x2)) = 2 + x_1 + x_2   
POL(e) = 2   
POL(DIV(x1, x2)) = 1 + (2)x_1 + (2)x_2   
POL(I(x1)) = 4 + (2)x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof

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

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 use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


DIV(div(X, Y), Z) → DIV(Y, div(i(X), Z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(i(x1)) = 3   
POL(div(x1, x2)) = 3 + x_2   
POL(e) = 3   
POL(DIV(x1, x2)) = (3)x_1   
The value of delta used in the strict ordering is 9.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.