Term Rewriting System R:
[x, y, z]
f(f(a, b), x) -> f(b, f(a, f(c, f(b, f(a, 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(b, f(a, f(c, f(b, f(a, x)))))
F(f(a, b), x) -> F(a, f(c, f(b, f(a, x))))
F(f(a, b), x) -> F(c, f(b, f(a, x)))
F(f(a, b), x) -> F(b, f(a, x))
F(f(a, b), 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
         ↳Negative Polynomial Order


Dependency Pairs:

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


Rules:


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





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

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


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

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


Used ordering:
Polynomial Order with Interpretation:

POL( F(x1, x2) ) = x1

POL( f(x1, x2) ) = x1

POL( a ) = 1

POL( b ) = 0

POL( c ) = 0


This results in one new DP problem. 


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




The following remains to be proven:
Dependency Pairs:

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


Rules:


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




The Proof could not be continued due to a Timeout.
Termination of R could not be shown.
Duration:
1:00 minutes