Factorize Boolean Function I have been wondering if this sum of products expression could be simplified to three terms. 
The whole function is 
$$yz + x'y + y'z' + xy' + xz$$
I have tried a few things but I could not work it out. Is there anything I could do about $yz + y'z'$?  
Thank you!
 A: The Consensus Theorem states that:
$$xy + x'z + yz = xy + x'z$$
Applied to your statement, we can therefore get rid of the $yz$ term because of the $x'y$ and $xz$ terms, and we can also get rid of the $xy'$ term because of the $xz$ term and the $y'z'$ terms. S, we are left with:
$$x'y+y'z'+xz$$
If you want to break this down to more elementary equivalence principles:
$$yz+x'y+y'z'+xy'+xz$$
$$\overset{Idempotence, Commmutation}{=}$$
$$yz+xz+x'y+xy'+xz+y'z'$$
$$\overset{Identity}{=}$$
$$yz1+xz+x'y+xy'1+xz+y'z'$$
$$\overset{Complement}{=}$$
$$yz(x+x')+xz+x'y+xy'(z + z')+xz+y'z'$$
$$\overset{Distribution}{=}$$
$$xyz+x'yz+xz+x'y+xy'z + xy'z'+xz+y'z'$$
$$\overset{Commutation}{=}$$
$$xyz+xz+x'yz+x'y+xy'z +xz+ xy'z'+y'z'$$
$$\overset{Distribution}{=}$$
$$xz(y+1)+x'y(z+1)+xz(y'+1)+y'z'(x+1)$$
$$\overset{Annihilation}{=}$$
$$xz1+x'y1+xz1+y'z'1$$
$$\overset{Identity}{=}$$
$$xz+x'y+xz+y'z'$$
$$\overset{Idempotence}{=}$$
$$xz+x'y+y'z'$$
A: $$yz + x'y + y'z' + xy' + xz = (yz′)′+(xy)′+xz$$
Indeed, using a truth table one can verify that: $(ab)' = ab' + a'b$
then it follows
$yz + y'z' = y''z+y'z'=(y'z)'$
$x'y + xy'=(xy)'$
thus the answer I stated as a comment.
