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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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


Rules:


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





The following dependency pairs can be strictly oriented:

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


The following rules can be oriented:

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


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

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


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




The following remains to be proven:
Dependency Pairs:

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


Rules:


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




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