Term Rewriting System R:
[x]
f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(s(s(x))) -> F(f(s(x)))
F(s(s(x))) -> F(s(x))

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

F(s(s(x))) -> F(s(x))
F(s(s(x))) -> F(f(s(x)))


Rules:


f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))





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

F(s(s(x))) -> F(f(s(x)))
two new Dependency Pairs are created:

F(s(s(0))) -> F(s(0))
F(s(s(s(x'')))) -> F(f(f(s(x''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Narrowing Transformation


Dependency Pairs:

F(s(s(s(x'')))) -> F(f(f(s(x''))))
F(s(s(x))) -> F(s(x))


Rules:


f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))





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

F(s(s(s(x'')))) -> F(f(f(s(x''))))
two new Dependency Pairs are created:

F(s(s(s(0)))) -> F(f(s(0)))
F(s(s(s(s(x'))))) -> F(f(f(f(s(x')))))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

F(s(s(s(s(x'))))) -> F(f(f(f(s(x')))))
F(s(s(s(0)))) -> F(f(s(0)))
F(s(s(x))) -> F(s(x))


Rules:


f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))





The following dependency pairs can be strictly oriented:

F(s(s(s(s(x'))))) -> F(f(f(f(s(x')))))
F(s(s(s(0)))) -> F(f(s(0)))
F(s(s(x))) -> F(s(x))


Additionally, the following rules can be oriented:

f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(s(x1))=  1 + x1  
  POL(f(x1))=  1  
  POL(F(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))





Using the Dependency Graph resulted in no new DP problems.

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