Term Rewriting System R:
[x]
f(a, f(f(a, x), a)) -> f(f(a, f(a, x)), a)

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(a, f(f(a, x), a)) -> F(f(a, f(a, x)), a)
F(a, f(f(a, x), a)) -> F(a, f(a, x))

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Semantic Labelling


Dependency Pair:

F(a, f(f(a, x), a)) -> F(a, f(a, x))


Rule:


f(a, f(f(a, x), a)) -> f(f(a, f(a, x)), a)





Using Semantic Labelling, the DP problem could be transformed. The following model was found:
f(x0, x1)=  0
a=  1
F(x0, x1)=  1

From the dependency graph we obtain 1 (labeled) SCCs which each result in correspondingDP problem.


   R
     ↳DPs
       →DP Problem 1
         ↳SemLab
           →DP Problem 2
             ↳Modular Removal of Rules


Dependency Pair:

F10(a, f01(f10(a, x), a)) -> F10(a, f10(a, x))


Rules:


f10(a, f01(f10(a, x), a)) -> f01(f10(a, f10(a, x)), a)
f10(a, f01(f11(a, x), a)) -> f01(f10(a, f11(a, x)), a)





We have the following set of usable rules:

f10(a, f01(f10(a, x), a)) -> f01(f10(a, f10(a, x)), a)
f10(a, f01(f11(a, x), a)) -> f01(f10(a, f11(a, x)), a)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(f_11(x1, x2))=  x1 + x2  
  POL(f_10(x1, x2))=  x1 + x2  
  POL(f_01(x1, x2))=  1 + x1 + x2  
  POL(F_10(x1, x2))=  x1 + x2  
  POL(a)=  0  

We have the following set D of usable symbols: {f11, f10, f01, F10, a}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

F10(a, f01(f10(a, x), a)) -> F10(a, f10(a, x))

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.


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