Proving $(xyz)' = x'+y'+z'$

Question

I'm trying to prove that $(xyz)' = x'+y'+z'$ using theorems/axioms.

I tried this but I just want to make sure if its the correct route or if I've done something "illegal"/wrong.

(xyz)' = [(xy)z]' by associativity 
       = [(x*y)'+z'] by DeMorgan's Law
       = [(x'+y') + z'] by DeMorgan's Law
       = [(x'+z')+(y'+z')] by Distribution
       = x'+y'+z' by simplifying redundant z' terms.

Is this the correct method?

Answer 1

I wouldn't say you did anything "illegal", but distribution is usually used in the following manner:

$\quad (x + y)z = xz + yz\quad $ or $\quad(xy)+ z = (x+z)(y+ z)$

or the "flip side"

$\quad x(y+z)=xy + xz \quad $ or $\quad x +(yz)=(x+ y)(x+z)$

Your work was done, essentially, when you after your second application of DeMorgan's. Then, we simply use associativity again:

$$[(x'+y') + z'] = x' + y' + z'\tag{by associativity}$$