Term Rewriting System R:
[x, y]
O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(O(x), O(y)) -> O'(+(x, y))
+'(O(x), O(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)
+'(I(x), O(y)) -> +'(x, y)
+'(I(x), I(y)) -> O'(+(+(x, y), I(0)))
+'(I(x), I(y)) -> +'(+(x, y), I(0))
+'(I(x), I(y)) -> +'(x, y)
*'(O(x), y) -> O'(*(x, y))
*'(O(x), y) -> *'(x, y)
*'(I(x), y) -> +'(O(*(x, y)), y)
*'(I(x), y) -> O'(*(x, y))
*'(I(x), y) -> *'(x, y)

Furthermore, R contains two SCCs. 


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


Dependency Pairs:

+'(I(x), I(y)) -> +'(x, y)
+'(I(x), I(y)) -> +'(+(x, y), I(0))
+'(I(x), O(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)
+'(O(x), O(y)) -> +'(x, y)


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)





The following dependency pairs can be strictly oriented:

+'(I(x), I(y)) -> +'(x, y)
+'(I(x), I(y)) -> +'(+(x, y), I(0))
+'(I(x), O(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)
+'(O(x), O(y)) -> +'(x, y)


The following usable rules using the Ce-refinement can be oriented:

+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
O(0) -> 0


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(I(x1))=  1 + x1  
  POL(0)=  0  
  POL(O(x1))=  1 + x1  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
+'(x1, x2) -> +'(x1, x2)
O(x1) -> O(x1)
I(x1) -> I(x1)
+(x1, x2) -> +(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS


Dependency Pair:


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

*'(I(x), y) -> *'(x, y)
*'(O(x), y) -> *'(x, y)


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)





The following dependency pair can be strictly oriented:

*'(I(x), y) -> *'(x, y)


There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(I(x1))=  1 + x1  
  POL(*'(x1, x2))=  x1 + x2  
  POL(O(x1))=  x1  

resulting in one new DP problem.
Used Argument Filtering System:
*'(x1, x2) -> *'(x1, x2)
I(x1) -> I(x1)
O(x1) -> O(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 4
             ↳Argument Filtering and Ordering


Dependency Pair:

*'(O(x), y) -> *'(x, y)


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)





The following dependency pair can be strictly oriented:

*'(O(x), y) -> *'(x, y)


There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(*'(x1, x2))=  x1 + x2  
  POL(O(x1))=  1 + x1  

resulting in one new DP problem.
Used Argument Filtering System:
*'(x1, x2) -> *'(x1, x2)
O(x1) -> O(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 4
             ↳AFS
             ...
               →DP Problem 5
                 ↳Dependency Graph


Dependency Pair:


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)





Using the Dependency Graph resulted in no new DP problems.

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