$$\overline{X}\overline{Y} + YZ + \overline{X}Y\overline{Z}$$
I'm having a lot of trouble simplifying this boolean expression.
I used commutative property and re-arranged it as my first step:
$$\overline{X}Y\overline{Z} + \overline{X}\overline{Y} + YZ$$
2nd Step: Factored out the common literal, $\overline{X}$.
$$\overline{X}(\overline{Y} + Y\overline{Z}) + YZ$$
3rd Step: ? I first used one of the identities to turn $\overline{Y} + Y$ into 1; I wasn't too sure what to do here and I believed what I did was wrong, so I then used the distributive property for the sum term in the parentheses instead. But, I don't know if this aforementioned step is correct.
Best Answer
$\overline{X}\overline{Y}+YZ+\overline{X}Y\overline{Z}=$
$\overline{X}\overline{Y}+\color{blue}{YZ}+\overline{X}Y\overline{Z}=$
$=\overline{X}\overline{Y}+\color{blue}{YZ\left(1+\overline{X}\right)}+\overline{X}Y\overline{Z}=$
$=\overline{X}\overline{Y}+YZ+\overline{X}YZ+\overline{X}Y\overline{Z}=$
$=\overline{X}\overline{Y}+YZ+\overline{X}Y\left(Z+\overline{Z}\right)=$
$=\overline{X}\overline{Y}+YZ+\overline{X}Y=$
$=\overline{X}\left(\overline{Y}+Y\right)+YZ=$
$=\overline{X}+YZ$
More details about the steps:
$\color{blue}{YZ=YZ\cdot1=YZ\left(1+\overline{X}\right)}$
because $\;1+\overline{X}=1\;$ whatever is the value of $\;X\;,$
indeed, if $\;X=0\;,\;$ then $\;1+\overline{X}=1+1=1\;,$
whereas if $\;X=1\;,\;$ then $\;1+\overline{X}=1+0=1\;.$