Term Rewriting System R:
[X, Y, Z]
div(X, e) -> i(X)
div(div(X, Y), Z) -> div(Y, div(i(X), Z))
i(div(X, Y)) -> div(Y, X)

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

DIV(X, e) -> I(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)
I(div(X, Y)) -> DIV(Y, X)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation


Dependency Pairs:

DIV(div(X, Y), Z) -> DIV(i(X), Z)
DIV(div(X, Y), Z) -> DIV(Y, div(i(X), Z))
I(div(X, Y)) -> DIV(Y, X)
DIV(X, e) -> I(X)


Rules:


div(X, e) -> i(X)
div(div(X, Y), Z) -> div(Y, div(i(X), Z))
i(div(X, Y)) -> div(Y, X)


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

DIV(X, e) -> I(X)
one new Dependency Pair is created:

DIV(div(X'', Y''), e) -> I(div(X'', Y''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳Narrowing Transformation


Dependency Pairs:

I(div(X, Y)) -> DIV(Y, X)
DIV(div(X'', Y''), e) -> I(div(X'', Y''))
DIV(div(X, Y), Z) -> DIV(Y, div(i(X), Z))
DIV(div(X, Y), Z) -> DIV(i(X), Z)


Rules:


div(X, e) -> i(X)
div(div(X, Y), Z) -> div(Y, div(i(X), Z))
i(div(X, Y)) -> div(Y, X)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

DIV(div(X, Y), Z) -> DIV(i(X), Z)
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 3
                 ↳Forward Instantiation Transformation


Dependency Pairs:

DIV(div(X'', Y''), e) -> I(div(X'', Y''))
DIV(div(X, Y), Z) -> DIV(Y, div(i(X), Z))
I(div(X, Y)) -> DIV(Y, X)


Rules:


div(X, e) -> i(X)
div(div(X, Y), Z) -> div(Y, div(i(X), Z))
i(div(X, Y)) -> div(Y, X)


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

I(div(X, Y)) -> DIV(Y, X)
two new Dependency Pairs are created:

I(div(X0, div(X'', Y''))) -> DIV(div(X'', Y''), X0)
I(div(e, div(X'''', Y''''))) -> DIV(div(X'''', Y''''), e)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 4
                 ↳Argument Filtering and Ordering


Dependency Pairs:

I(div(e, div(X'''', Y''''))) -> DIV(div(X'''', Y''''), e)
DIV(div(X, Y), Z) -> DIV(Y, div(i(X), Z))
I(div(X0, div(X'', Y''))) -> DIV(div(X'', Y''), X0)
DIV(div(X'', Y''), e) -> I(div(X'', Y''))


Rules:


div(X, e) -> i(X)
div(div(X, Y), Z) -> div(Y, div(i(X), Z))
i(div(X, Y)) -> div(Y, X)


Strategy:

innermost




The following dependency pairs can be strictly oriented:

I(div(e, div(X'''', Y''''))) -> DIV(div(X'''', Y''''), e)
DIV(div(X, Y), Z) -> DIV(Y, div(i(X), Z))
I(div(X0, div(X'', Y''))) -> DIV(div(X'', Y''), X0)
DIV(div(X'', Y''), e) -> I(div(X'', Y''))


The following usable rules for innermost can be oriented:

div(X, e) -> i(X)
div(div(X, Y), Z) -> div(Y, div(i(X), Z))
i(div(X, Y)) -> div(Y, X)


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{I, DIV} > {i, div}

resulting in one new DP problem.
Used Argument Filtering System:
DIV(x1, x2) -> DIV(x1, x2)
div(x1, x2) -> div(x1, x2)
i(x1) -> i(x1)
I(x1) -> I(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 5
                 ↳Dependency Graph


Dependency Pair:


Rules:


div(X, e) -> i(X)
div(div(X, Y), Z) -> div(Y, div(i(X), Z))
i(div(X, Y)) -> div(Y, X)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes