[Math] Boolean Algebra: Simplifying $\;xyz + x’y + xyz’$

boolean-algebralogic

Given the following expression: $xyz + x'y + xyz'\,$ where ($'$) means complement, I tried to simplify it by first factoring out a y so I would get $\;y(xz + x' + xz').\,$

At this point, it appears I have several options:

A) Use two successive rounds of distributive property:

$\begin{align} y( (x + x')(z + x') + xz') )
&= y ( z + x' + xz')\\ & = y ( z + (x' + x)(x' + z') )\\ &= y ( z + x' + z') \\ &= y ( x') \\ &= yx'\end{align}$

B) Or I could use absorption,

$\begin{align}y ( xz + xz' + x' )
&= y ( x (z+z') + x') \\
& = y ( x + x' )\\
&= y ( 1) \\
&= y\end{align}$

I believe the second answer is correct. What am I doing wrong with option A ?

Best Answer

Using the distributive property (first method), we get:

$$\begin{align} xyz + x'y + xyz' & = xy(z + z') + x'y \\ &= xy + x'y \\&= (x + x')y \\&= y\end{align}$$

You erred when you went from $ y ( z + x' + z') $ to $yx'$. You should have $$y((z+z')+x')= y(1+x') = y\cdot 1 = y$$

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