Basic prove that boolean function is self-dual

Question

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 :).

Answer 1

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.

Answer 2

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 $f(x,y,z)$.

Answer 3

$$f(x,y,z) = \overline x\ \overline y\ \overline z + \overline x yz +xy \overline z+x\overline y z$$

$$f_1^D(x,y,z) = \overline {x y z + x \ \overline y\ \overline z + \overline x\ \overline y z + \overline x\ y\ \overline z}$$

Take DeMorgan's. $$ \overline {x y z} \cdot \overline {x \ \overline y\ \overline z} \cdot \overline {\overline x\ \overline y z} \cdot \overline {\overline x\ y\ \overline z}$$

Take DeMorgan's on terms. $$ (\overline x + \overline y + \overline z) \cdot (\overline x + y + z) \cdot ( x + y + \overline z) \cdot (x + \overline y + z)$$

Simplify. $$ (\overline x + \overline y + \overline z) \cdot (\overline x + y + z)$$ $$ \overline x\ \overline x + \overline x y + \overline x z + \overline y\ \overline x + \overline y y + \overline y z + \overline z \ \overline x + \overline z y + \overline z z$$ $$ \overline x + \overline x y + \overline x z + \overline y\ \overline x + 0 + \overline y z + \overline z \ \overline x + \overline z y + 0 $$ $$ \overline x (1 + y + z + \overline y\ + \overline z) + \overline y z + \overline z y $$ $$ \overline x + \overline y z + \overline z y $$

The remainder.

$$ ( x + y + \overline z) \cdot (x + \overline y + z)$$ $$ xx + x\overline y + xz + yx + y\overline y + yz + \overline z x + \overline z\ \overline y + \overline z z$$ $$ x + yz + \overline z\ \overline y$$

Put it together. $$ (\overline x + \overline y z + \overline z y) \cdot (x + yz + \overline z\ \overline y)$$ $$ \overline x x + \overline x yz + \overline x\ \overline z\ \overline y + \overline y z x + \overline y z yz + \overline y z \overline z\ \overline y + \overline z y x + \overline z y yz + \overline z y \overline z\ \overline y$$ $$ 0 + \overline x yz + \overline x\ \overline z\ \overline y + \overline y z x + 0 + 0 + \overline z y x + 0 + 0$$ $$\overline x\ \overline y\ \overline z + \overline x y z + x y \overline z + x \overline y z$$

$f(x,y,z) = f_1^D(x,y,z) $, so self-dual.