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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


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


Dependency Pairs:

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


Rules:


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





The following dependency pairs can be strictly oriented:

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


There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(s(x1))=  x1  
  POL(F(x1))=  x1  
  POL(+(x1, x2))=  1 + x1 + x2  

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


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


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.

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