Boolean Algebra: Simplifying $\;xyz + x'y + xyz'$
Matthew Harrington 
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 ?
$\endgroup$4 Answers
$\begingroup$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 $$y((z+z')+x')= y(1+x') = y\cdot 1 = y$$
$\endgroup$ $\begingroup$For option A, you made the error of stating $x' +1 = x'$ when its really $x' + 1 = 1$. Hence you would get the same answer for both options.
$\endgroup$ 1 $\begingroup$Second answer is correct as can also be done as follows \begin{align} xyz+x^{\prime}y+xyz^{\prime}&=xyz+xyz^{\prime}+x^{\prime}y\qquad\text{commutative law}\\&=xy(z+z^{\prime})+x^{\prime}y\\&=xy.1+x^{\prime}y\\&=(x+x^{\prime})y=1.y=y \end{align} While as in A) Note that \begin{align} y((x+x^{\prime})(z+x^{\prime})+xz^{\prime})&=y(z+x^{\prime}+xz^{\prime})\\&=yz+x^{\prime}y+xyz^{\prime}\neq xyz+x^{\prime}y+xyz^{\prime} \end{align}
$\endgroup$ $\begingroup$xyz + x'y + xyz'=
multiplying (z+z') in x'y = x'y(z+z')=x'y
z+z'=1 (do a truth table)
xyz + x'y(z+z') + xyz'
xyz + (x'yz + x'yz') + xyz'
take common
(xyz + x'yz) +( x'yz' + xyz')
{((yz(x+x')) + (yz'(x+x')) }
yz+yz'
y(z+z')
:Y
$\endgroup$ 1
