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)

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
         ↳Argument Filtering and Ordering


Dependency Pairs:

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))
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)





The following dependency pairs can be strictly oriented:

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


The following usable rules w.r.t. to the AFS can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(I(x1))=  x1  
  POL(i(x1))=  x1  
  POL(e)=  0  
  POL(DIV(x1, x2))=  x1 + x2  
  POL(div(x1, x2))=  1 + x1 + x2  

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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 2
             ↳Dependency Graph


Dependency Pairs:

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





Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 3
                 ↳Argument Filtering and Ordering


Dependency Pair:

DIV(div(X, Y), Z) -> DIV(Y, 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)





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(div(x1, x2))=  1 + x1 + x2  

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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 4
                 ↳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)





Using the Dependency Graph resulted in no new DP problems.

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