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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Semantic Labelling


Dependency Pairs:

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


Rules:


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





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

F00(y, f10(x, f11(a, x))) -> F00(f10(a, f11(x, a)), f10(a, y))
F00(y, f00(x, f10(a, x))) -> F00(f10(a, f01(x, a)), f10(a, y))


Rules:


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





We have the following set of usable rules:

f10(x, f11(x, y)) -> f01(f01(f11(x, a), a), a)
f10(y, f00(x, f10(a, x))) -> f00(f10(a, f01(x, a)), f11(a, y))
f10(y, f10(x, f11(a, x))) -> f00(f10(a, f11(x, a)), f11(a, y))
f10(x, f10(x, y)) -> f01(f01(f11(x, a), a), 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_00(x1, x2))=  x1 + x2  
  POL(f_00(x1, x2))=  x1 + x2  
  POL(f_10(x1, x2))=  x1 + x2  
  POL(f_01(x1, x2))=  x1 + x2  
  POL(a)=  0  

We have the following set D of usable symbols: {f11, F00, f00, f10, f01, a}
No Dependency Pairs can be deleted.
4 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(y, f10(x, f11(a, x))) -> F00(f10(a, f11(x, a)), f10(a, y))


Rules:


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





Removed all semantic labels.

   R
     ↳DPs
       →DP Problem 1
         ↳SemLab
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 4
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

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


Rules:


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




Termination of R could not be shown.
Duration:
0:09 minutes