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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


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





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

F(a(x), a(y)) -> A(f(x, y))
two new Dependency Pairs are created:

F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(a(b(x'')), a(b(y''))) -> A(b(f(x'', y'')))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

A(a(f(x, y))) -> A(y)
A(a(f(x, y))) -> A(x)
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(a(x), a(y)) -> F(x, y)
F(b(x), b(y)) -> F(x, y)


Rules:


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





The following dependency pairs can be strictly oriented:

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


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

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


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

A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(a(x), a(y)) -> F(x, y)
F(b(x), b(y)) -> F(x, y)


Rules:


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




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