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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Semantic Labelling


Dependency Pair:

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


Rule:


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





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

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:

F00(f00(y, z), f00(x, f00(a, x))) -> F00(f00(f00(a, z), f00(x, a)), f00(a, y))


Rules:


f00(f00(y, z), f00(x, f00(a, x))) -> f00(f00(f00(a, z), f00(x, a)), f00(a, y))
f00(f00(y, z), f10(x, f01(a, x))) -> f00(f00(f00(a, z), f10(x, a)), f00(a, y))
f00(f01(y, z), f00(x, f00(a, x))) -> f00(f00(f01(a, z), f00(x, a)), f00(a, y))
f00(f01(y, z), f10(x, f01(a, x))) -> f00(f00(f01(a, z), f10(x, a)), f00(a, y))
f00(f10(y, z), f00(x, f00(a, x))) -> f00(f00(f00(a, z), f00(x, a)), f01(a, y))
f00(f10(y, z), f10(x, f01(a, x))) -> f00(f00(f00(a, z), f10(x, a)), f01(a, y))
f00(f11(y, z), f00(x, f00(a, x))) -> f00(f00(f01(a, z), f00(x, a)), f01(a, y))
f00(f11(y, z), f10(x, f01(a, x))) -> f00(f00(f01(a, z), f10(x, a)), f01(a, y))





We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(F_00(x1, x2))=  1 + x1 + x2  
  POL(f_00(x1, x2))=  x1 + x2  
  POL(a)=  0  

We have the following set D of usable symbols: {F00, f00, a}
No Dependency Pairs can be deleted.
8 non usable rules have been deleted.

The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳SemLab
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 3
                 ↳Unlabel


Dependency Pair:

F00(f00(y, z), f00(x, f00(a, x))) -> F00(f00(f00(a, z), f00(x, a)), f00(a, y))


Rule:

none





Removed all semantic labels.

   R
     ↳DPs
       →DP Problem 1
         ↳SemLab
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 4
                 ↳Semantic Labelling


Dependency Pair:

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


Rule:

none





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

From the dependency graph we obtain 0 (labeled) SCCs which each result in correspondingDP problem.

Termination of R successfully shown.
Duration:
0:01 minutes