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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Semantic Labelling


Dependency Pair:

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


Rule:


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





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 Pairs:

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


Rules:


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





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_11(x1, x2))=  x1 + x2  
  POL(F_00(x1, x2))=  1 + 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 Pairs:

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


Rule:

none





Removed all semantic labels.

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


Dependency Pair:

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


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 1 (labeled) SCCs which each result in correspondingDP problem.


   R
     ↳DPs
       →DP Problem 1
         ↳SemLab
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 5
                 ↳Modular Removal of Rules


Dependency Pair:

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


Rule:

none





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_11(x1, x2))=  x1 + x2  
  POL(F_00(x1, x2))=  1 + 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, a}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

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