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
         ↳Argument Filtering and 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 pair can be strictly oriented:

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


The following usable rules for innermost 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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
a > b

resulting in one new DP problem.
Used Argument Filtering System:
F(x1, x2) -> x1
f(x1, x2) -> x1


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




The following remains to be proven:
Dependency Pairs:

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