Factorize Boolean Function

Question

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!

Answer 1

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:

$$yz+x'y+y'z'+xy'+xz$$

$$\overset{Idempotence, Commmutation}{=}$$

$$yz+xz+x'y+xy'+xz+y'z'$$

$$\overset{Identity}{=}$$

$$yz1+xz+x'y+xy'1+xz+y'z'$$

$$\overset{Complement}{=}$$

$$yz(x+x')+xz+x'y+xy'(z + z')+xz+y'z'$$

$$\overset{Distribution}{=}$$

$$xyz+x'yz+xz+x'y+xy'z + xy'z'+xz+y'z'$$

$$\overset{Commutation}{=}$$

$$xyz+xz+x'yz+x'y+xy'z +xz+ xy'z'+y'z'$$

$$\overset{Distribution}{=}$$

$$xz(y+1)+x'y(z+1)+xz(y'+1)+y'z'(x+1)$$

$$\overset{Annihilation}{=}$$

$$xz1+x'y1+xz1+y'z'1$$

$$\overset{Identity}{=}$$

$$xz+x'y+xz+y'z'$$

$$\overset{Idempotence}{=}$$

$$xz+x'y+y'z'$$

Answer 2

$$yz + x'y + y'z' + xy' + xz = (yz′)′+(xy)′+xz$$

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.