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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Semantic Labelling


Dependency Pairs:

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


Rule:


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


Strategy:

innermost




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:

F01(f10(a, x), y) -> F00(x, f11(a, y))
F00(f11(a, x), y) -> F01(f10(x, f10(a, y)), a)
F00(f10(a, x), y) -> F00(x, f10(a, y))


Rules:


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


Strategy:

innermost




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))=  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, 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(f11(a, x), y) -> F01(f10(x, f10(a, y)), a)
F00(f10(a, x), y) -> F00(x, f10(a, y))
F01(f10(a, x), y) -> F00(x, f11(a, y))


Rule:

none


Strategy:

innermost




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 Pairs:

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


Rule:

none


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:02 minutes