I'm fairly new too boolean algebra. I've tried simplifying this equation but I'm not quite sure if I've done it correctly.
Simplify to 1 literal, (X + Y + Y'Z)(Y + X)(Y + X')
My attempt:
(X+Y+Y'Z)(Y+X)(Y+X')
[(X+Y+Y'Z) + (Y+X) + (Y+X')]'
[X + Y + Y'Z + Y+ X + Y + X']'
[(X+X) + (Y+Y+Y) + Y'Z + X']'
[X + Y + Y'Z + X']'
[(X+X') + Y + Y'Z]'
[1 + Y + Y'Z]'
[1]'
0
I'm also having trouble simplifying this equation to 3 literals AB'+ AC + C'B
Any help is appreciated.
On
You misapplied DeMorgan's at your very first step. Let's start from the "get-go."
$$(X + Y + Y'Z)\underbrace{(Y + X)(Y + X')}_{\large = Y + (\underbrace{XX'}_{\large \, 0}) = Y} $$ This gives us $$(X + Y + Y'Z)(Y) = XY + \underbrace{YY}_{\large =\, Y} + \underbrace{\underbrace{YY'}_{\large=\, 0}Z}_{\large =\, 0} = XY + Y = \underbrace{(X + 1)}_{\large = \,1}Y = Y$$
Alternatively, we can use the distributive law and note that $$(X+Y+Y'Z)(Y+X)(Y+X') = Y+(\underbrace{(X + Y'Z)\underbrace{(X)(X')}_{\large = \,0}}_{\large = \,0} = Y$$
One of the reason we talk about boolean algebra is that you have an algebraic structure that allows you to manipulate symbols the same way you do for any variable. So instead of attempting a 3 part DeMorgan's law, try using the distributive property.
$(Y+X)(Y+X') = YY+X'Y+XY+XX' = Y + (X'+X)Y + 0 = Y + (1)Y = Y$
The biggest difference from regular algebra are the extra boolean laws you need to follow. The following list is not exhaustive but gives enough to start manipulating. Remember that we use multiplication (juxtaposition) to represent boolean AND and addition to represent boolean OR.
$XX' = 0 \\ X+X' = 1 \\ X+X = X \\ XX = X \\ XY+X = X$
The final expression I got for the first question turns out to be $Y$. See if you can finish the details.