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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(s(x), s(y)) -> +'(s(x), +(y, 0))
+'(s(x), s(y)) -> +'(y, 0)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pair:

+'(s(x), s(y)) -> +'(s(x), +(y, 0))


Rules:


+(0, y) -> y
+(s(x), 0) -> s(x)
+(s(x), s(y)) -> s(+(s(x), +(y, 0)))


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

+'(s(x), s(y)) -> +'(s(x), +(y, 0))
two new Dependency Pairs are created:

+'(s(x), s(0)) -> +'(s(x), 0)
+'(s(x), s(s(x''))) -> +'(s(x), s(x''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Forward Instantiation Transformation


Dependency Pair:

+'(s(x), s(s(x''))) -> +'(s(x), s(x''))


Rules:


+(0, y) -> y
+(s(x), 0) -> s(x)
+(s(x), s(y)) -> s(+(s(x), +(y, 0)))


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

+'(s(x), s(s(x''))) -> +'(s(x), s(x''))
one new Dependency Pair is created:

+'(s(x0), s(s(s(x'''')))) -> +'(s(x0), s(s(x'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 3
                 ↳Polynomial Ordering


Dependency Pair:

+'(s(x0), s(s(s(x'''')))) -> +'(s(x0), s(s(x'''')))


Rules:


+(0, y) -> y
+(s(x), 0) -> s(x)
+(s(x), s(y)) -> s(+(s(x), +(y, 0)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

+'(s(x0), s(s(s(x'''')))) -> +'(s(x0), s(s(x'''')))


There are no usable rules for innermost that need to be oriented.

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Dependency Graph


Dependency Pair:


Rules:


+(0, y) -> y
+(s(x), 0) -> s(x)
+(s(x), s(y)) -> s(+(s(x), +(y, 0)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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