Term Rewriting System R:
[x, y, z]
i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

I(+(x, y)) -> +'(i(x), i(y))
I(+(x, y)) -> I(x)
I(+(x, y)) -> I(y)
+'(x, +(y, z)) -> +'(+(x, y), z)
+'(x, +(y, z)) -> +'(x, y)

Furthermore, R contains two SCCs. 


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


Dependency Pairs:

+'(x, +(y, z)) -> +'(x, y)
+'(x, +(y, z)) -> +'(+(x, y), z)


Rules:


i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x





The following dependency pair can be strictly oriented:

+'(x, +(y, z)) -> +'(x, y)


The following rules can be oriented:

+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x
i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))


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

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


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


Dependency Pair:

+'(x, +(y, z)) -> +'(+(x, y), z)


Rules:


i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x





The following dependency pair can be strictly oriented:

+'(x, +(y, z)) -> +'(+(x, y), z)


The following rules can be oriented:

+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x
i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))


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

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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳AFS
             ...
               →DP Problem 4
                 ↳Dependency Graph
       →DP Problem 2
         ↳AFS


Dependency Pair:


Rules:


i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x





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


Rules:


i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x





The following dependency pairs can be strictly oriented:

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


The following rules can be oriented:

i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x


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

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


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


Dependency Pair:


Rules:


i(0) -> 0
i(i(x)) -> x
i(+(x, y)) -> +(i(x), i(y))
+(0, y) -> y
+(x, 0) -> x
+(i(x), x) -> 0
+(x, i(x)) -> 0
+(x, +(y, z)) -> +(+(x, y), z)
+(+(x, i(y)), y) -> x
+(+(x, y), i(y)) -> x





Using the Dependency Graph resulted in no new DP problems.

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