What is the most simplified form of $y(x′z + xz′) + x(yz + yz′)$

Question

I am stuck on a problem that I know the logical answer to, yet I cannot seem to simplify properly to get there.

• I want to simplify $$F(x,y,z)=y(x′z + xz′) + x(yz + yz′)$$
• I know the simplest form (using a truth table) is $$xy + yz$$
• Yet my simplification yields $$xy + x’yz$$

My steps are below with the identities used

$y(x′z + xz′) + x(yz + yz′)$
$y(x′z + xz′) + x(y(z + z’))$ Distributive
$y(x′z + xz′) + x(y(1))$ Inverse
$y(x′z + xz′) + xy$ Identity
$yx’z + yxz’ + xy$ Distributive
$y(x’z + xz’ + x)$ Distributive
$y(x’z + x)$ Absoption
$yx’z + yx$ Distributive
$\mathbf {xy + x’yz}$ Commutative x 2

Where did I go wrong?

Answer 1

Continuing from your simplification:

$$\begin{array}{ccl} xy+x'yz&=&xy(z+z')+x'yz\\ &=& xyz+xyz'+x'yz\\ &=& xyz+xyz+xyz'+x'yz\\ &=& (xyz+xyz')+(xyz+x'yz)\\ &=& xy+yz \end{array}$$

Answer 2

$x+z$ is equivalent to $x+x'z$. I don't remember the law. Absorption may be? The proof can be obtained from the true/false table or as follows: $$ (x+x'z)'=x' (x+z')=x'z'=(x+z)' $$

Answer 3

I would probably just write $$ \begin{eqnarray} y(x′z+xz′)+x(yz+yz′) &=& y(x'z+xz')+xyz+xyz' \\ &=&y(x'z+xz')+yxz+yxz' \\ &=&y(x'z+xz'+xz+xz') \\ &=&y(x'z+xz+xz') \\ &=&y(x'z+xz+xz+xz') \\ &=&y((x'+x)z+x(z+z')) \\ &=&y(z+x). \end{eqnarray} $$