Basic prove that boolean function is self-dual I'm tring to prove this function: 
   $$ f(x,y,z) = x'y'z'+x'yz+xyz'+xy'z $$

is self-dual, I've tried some basic manipulations like using double not on the function with de-morgan rules but got no results.
I also looked at Kerno's map and it's really suppose to be self-dual.
I would really appreciate if someone would guide me how to prove this basic thing...

thanks :).
Since $f(x,y,z)$ is true if and only if an even number (2 or 0) of $x,y$, and $z$ are true, the dual of $f(x,y,z)$ will be true if and only if it is not the case that an even number of $\neg x,\neg y$, and $\neg z$ are true, i.e., if and only if an odd number (1 or 3) of $x,y$, and $z$ are false. But that's equivalent to the original criterion for $f(x,y,z)$ being true, so the dual of $f(x,y,z)$ is the same Boolean function as $f(x,y,z)$.
See here, a function is self-dual if it is equal to its own dual. 
So you simply take the dual of this function, and if it gives you the original function, then its self-dual. Otherwise, its not.

For what its worth, I did the calculation, and I didn't find it to be dual.