Term Rewriting System R:
[x, y, z]
f(f(f(a, b), c), x) -> f(b, f(a, f(c, 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(f(a, b), c), x) -> F(b, f(a, f(c, f(b, x))))
F(f(f(a, b), c), x) -> F(a, f(c, f(b, x)))
F(f(f(a, b), c), x) -> F(c, f(b, x))
F(f(f(a, b), c), 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
         ↳Modular Removal of Rules


Dependency Pairs:

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


Rules:


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





We have the following set of usable rules:

f(x, f(y, z)) -> f(f(x, y), z)
f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(c)=  0  
  POL(b)=  0  
  POL(a)=  1  
  POL(F(x1, x2))=  1 + x1 + x2  
  POL(f(x1, x2))=  x1 + x2  

We have the following set D of usable symbols: {c, b, a, F, f}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

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

No Rules can be deleted.

The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳Negative Polynomial Order


Dependency Pairs:

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


Rules:


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





The following Dependency Pair can be strictly oriented using the given order.

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


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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


Used ordering:
Polynomial Order with Interpretation:

POL( F(x1, x2) ) = x1

POL( f(x1, x2) ) = x1

POL( a ) = 1

POL( b ) = 0


This results in one new DP problem. 


   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳Neg POLO
             ...
               →DP Problem 3
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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


Rules:


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




Termination of R could not be shown.
Duration:
0:41 minutes