Using DeMorgan's Laws to complement a function Using DeMorgan's Law, write an expression for the complement of $F$ if: 


*

*$F(x,y,z) = x(y' + z)$.
$F=x'+(y'+x)'$

*$F(x,y,z) = xy + x'z + yz'$
$F=(xy)'(x'z)'(yz')'$

*$F(w,x,y,z) = xyz' (y'z + x)' + (w'yz + x' )$.
$F=[(xyz')'+(y'z+x)](w'yz+x')'$
My answers are underneath the numbered questions. Is everything correct? I'm not 100% sure as to what I'm exactly supposed to do. I just took all the ANDs, negated them and made them ORs and vice-versa.
 A: You don't have to stop there. Continue using De Morgan until you get to these simpler forms.


*

*$F^c = x' + y z'$

*$F^c = (x'+y')(x + z')(y'+z)$

*$F^c = x(w+y' +z')$

Since you asked, let me briefly explain how I got (3). At some point during the reduction you will come across a long formula like this:
$((x' + y' + z) + (y'z + x))((w + y' + z')x)$
Now, noticing that disjunction (+) is associative we can collapse the parens on the left to get:
$(x' + y' + z + y'z + x)((w + y' + z')x)$
And here we immediately realize that we're dealing with the tautology $(x + x')$:
$(\color{blue}{x' +} y' + z + y'z + \color{blue}{x})((w + y' + z')x)$
This means that the entire formula on the left is materially equivalent ($\equiv$) to $\top$:
$\color{blue}{\top}((w + y' + z')x)$
Now since $(\top \land \phi) \equiv \phi$, for all $\phi$, we cancel it, getting:
$(w + y' + z')x$
Lastly, since conjunction is commutative, I just put the x in front to obtain:
$x(w + y' + z')$
A: To be clear about the rules of logic, Demorgan's Laws dictate the following:
$(ab)' = a' + b' \\
(a+b)' = a'b'$
The mnemonic that helps me remember Demorgan's Laws: "Break the negation/complement line, change the sign." Note that a complement of a complement (double apostrophes) is just the original argument. I.e., $(a')' = a'' = a$


*

*$F = x(y'+z) \\ F' = [x(y'+z)]' \\ = x' + [y'+z]' \\ = x'+ y''z' \\ = x'+yz'$

*$F = xy+x'z+yz' \\ F'= [xy+x'z+yz']' = [(xy+x'z)+(yz')]' = (xy+x'z)'(yz')' \\ = [(xy)+(x'z)]'[y'+z'']  = (xy)'(x'z)'[y'+z]  =(x'+y')(x''+z')(y'+z) \\ = (x'+y')(x+z')(y'+z) \\ = [x'x+x'z'+xy'+y'z'](y'+z) \\ =[0+x'z'+xy'+y'z'](y'+z) \\ = x'y'z'+x'z'z+xy'y'+xy'z+y'y'z'+y'z'z \\ = x'y'z'+x'(0)+x(y')+xy'z+(y')z'+y'(0) \\ = x'y'z'+xy'+xy'z+y'z' \\ F' = y'(x'z'+x+xz+z')$

*$F = xyz' (y'z + x)' + (w'yz + x' ) \\ F' = [xyz' (y'z + x)' + (w'yz + x' )]' = [xyz'(y'z+x)']'[(w'yz+x')]' \\ = [[xyz']'+[x+y'z]'][[w'yz]'[x']'] = [(x'+y'+z'')+x'(y'z')'][(w''+y'+z')x'' \\ = [x'+y'+z+x'(y''+z'')][x(w+y'+z')]\\ = x(x'+y'+z+x'y+x'z)(w+y'+z') \\ = (xx'+xy'+xz+xx'y+xx'z)(w+y'+z') \\ = (0+xy'+xz+0)(w+y'+z') \\ = wxy'+xy'y'+xy'z'+wxz+xy'z+xzz' \\ = wxy'+x(1)+xy'z'+wxz+xy'z+x(0) \\ F'= wxy'+x+xy'z'+wxz+xy'z$
