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
         ↳Modular Removal of Rules


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




We have the following set of usable rules:

div(div(X, Y), Z) -> div(Y, div(i(X), Z))
div(X, e) -> i(X)
i(div(X, Y)) -> div(Y, X)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(I(x1))=  x1  
  POL(i(x1))=  x1  
  POL(e)=  0  
  POL(DIV(x1, x2))=  x1 + x2  
  POL(div(x1, x2))=  x1 + x2  

We have the following set D of usable symbols: {I, i, DIV, div}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

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

The following rules can be deleted as they contain symbols in their lhs which do not occur in D:

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


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳Modular Removal of Rules


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)


Rules:


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


Strategy:

innermost




We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(I(x1))=  x1  
  POL(i(x1))=  x1  
  POL(DIV(x1, x2))=  x1 + x2  
  POL(div(x1, x2))=  x1 + x2  

We have the following set D of usable symbols: {i, DIV, div}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

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

2 non usable rules have been deleted.

The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 3
                 ↳Modular Removal of Rules


Dependency Pairs:

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


Rule:

none


Strategy:

innermost




We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(i(x1))=  x1  
  POL(DIV(x1, x2))=  1 + x1 + x2  
  POL(div(x1, x2))=  1 + x1 + x2  

We have the following set D of usable symbols: {i, DIV, div}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

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

No Rules can be deleted.

The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 4
                 ↳Size-Change Principle


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. DIV(div(X, Y), Z) -> DIV(Y, div(i(X), Z))
and get the following Size-Change Graph(s):
{1} , {1}
1>1

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
i(x1) -> i(x1)
div(x1, x2) -> div(x1, x2)

We obtain no new DP problems.

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