# Trying to simplify complicated boolean expression

$$U=(X+\overline{Y}+\overline{W}Z)(WY+Y+\overline{W}Z)$$

I have been trying to simplify the above expression for over an hour and can't figure out how to go about it. The results that I have gotten have been Z and just 1. I'm not using the rules correctly or something. I feel like if someone can get me started on the right track then I can do the rest.

Thanks for any help.

EDIT: After FOIL and simplifying the easy terms I'm left with this: $$U=XWY+XY+X\overline{W}Z+\overline{Y}\overline{W}Z+\overline{W}ZY+\overline{W}Z$$ Should I take out like terms or is there something else I can use? Is this allowed: $$U=X(WY+Y+\overline{W}Z)+\overline{W}Z(\overline{Y}+Y+1)$$Again thanks for any help.

• So what rules are you using and how are you using them? – Brian Drummond Feb 11 '18 at 20:02
• @BrianDrummond The first rule I am using is the distributive rule, so that means I kind of FOIL them. After that, there are some terms that come out to zero and one term is AND with itself so it equals itself. At that point, I have a very long expression that consists of many terms. I tried pulling out some of the like terms and simplifying from there, but that is how I got Z. If you need me to, I can write out the steps I took, but it is very long and I would like to avoid that if possible. Thank you. – JustHeavy Feb 11 '18 at 20:05
• No Karnaugh map? – Brian Drummond Feb 11 '18 at 20:29
• imagine you are playing a game where you drive around trying to find things on a list that is in a form of a boolean expression win=(blue house)(red door)+(blue house)(green window)+(blue house) .... what would you look for? – jsotola Feb 11 '18 at 20:37
• Any logical formula can be reduced using Karnaugh Maps... That is literally the solution to every single logical reduction problem you can ever think of... Except perhaps, if you're trying to create some delay by adding more gates. – KingDuken Feb 11 '18 at 23:01

Applying the distributive rule is the correct approach.

(X + Y' + W'Z)(WY + Y + W'Z)

But First simplify WY + Y to just Y.
(X + Y' + W'Z)(Y + W'Z)

Now distribute...
Y(X + Y' + W'Z) + W'Z(X + Y' + W'Z) YX + YY' + YW'Z + W'ZX + W'ZY' + W'ZW'Z

Note that the YY' term disappears. W'ZW'Z becomes just W'Z.
YX + YW'Z + W'ZX + W'ZY' + W'Z

Next note that YW'Z + Y'W'Z simplifies to W'Z.
YX + W'Z + W'ZX + W'Z

Note that W'Z + W'ZX + W'Z is just W'Z. So finally we have...
YX + W'Z