Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P 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))
DIV(div(X, Y), Z) → DIV(i(X), Z)
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.

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DIV(X, e) → I(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(DIV(x₁, x₂)) = x₁ + x₂   
POL(I(x₁)) = x₁   
POL(div(x₁, x₂)) = x₁ + x₂   
POL(e) = 1   
POL(i(x₁)) = x₁   

The following usable rules [FROCOS05] were oriented:

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

(4) Obligation:

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

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(DIV(x₁, x₂)) = x₁   
POL(I(x₁)) = x₁   
POL(div(x₁, x₂)) = 1 + x₁ + x₂   
POL(e) = 0   
POL(i(x₁)) = x₁   

The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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.

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) TRUE