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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Polynomial interpretation [POLO]:

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

From the DPs we obtained the following set of size-change graphs:

• I(div(X, Y)) → DIV(Y, X) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1, 1 > 2

• DIV(div(X, Y), Z) → DIV(i(X), Z) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1, 2 >= 2

• DIV(div(X, Y), Z) → DIV(Y, div(i(X), Z)) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• DIV(div(X, Y), Z) → I(X) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• DIV(X, e) → I(X) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 >= 1

We oriented the following set of usable rules [AAECC05,FROCOS05].

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

(4) TRUE