Term Rewriting System R:
[x]
sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

SUM(s(x)) -> SUM(x)
SUM1(s(x)) -> SUM1(x)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

SUM(s(x)) -> SUM(x)


Rules:


sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))





The following dependency pair can be strictly oriented:

SUM(s(x)) -> SUM(x)


The following rules can be oriented:

sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(SUM(x1))=  x1  
  POL(0)=  0  
  POL(sum1(x1))=  x1  
  POL(s(x1))=  1 + x1  
  POL(sum(x1))=  x1  

resulting in one new DP problem.
Used Argument Filtering System:
SUM(x1) -> SUM(x1)
s(x1) -> s(x1)
sum(x1) -> sum(x1)
+(x1, x2) -> x1
sum1(x1) -> sum1(x1)


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


Dependency Pair:


Rules:


sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, 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 Pair:

SUM1(s(x)) -> SUM1(x)


Rules:


sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))





The following dependency pair can be strictly oriented:

SUM1(s(x)) -> SUM1(x)


The following rules can be oriented:

sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(sum1(x1))=  x1  
  POL(s(x1))=  1 + x1  
  POL(sum(x1))=  x1  
  POL(SUM1(x1))=  x1  

resulting in one new DP problem.
Used Argument Filtering System:
SUM1(x1) -> SUM1(x1)
s(x1) -> s(x1)
sum(x1) -> sum(x1)
+(x1, x2) -> x1
sum1(x1) -> sum1(x1)


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


Dependency Pair:


Rules:


sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))





Using the Dependency Graph resulted in no new DP problems.

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