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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


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

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


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

resulting in one new DP problem. 



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




The following remains to be proven:
Dependency Pairs:

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


Rules:


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


Strategy:

innermost



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