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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

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


Rule:


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


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) -> F(a, f(b, x))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Polynomial Ordering


Dependency Pairs:

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


Rule:


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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


Additionally, the following usable rule for innermost can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(b)=  0  
  POL(a)=  1  
  POL(f(x1, x2))=  x1  
  POL(F(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 3
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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


Rule:


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


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:00 minutes