#### wilmania

Joined Nov 7, 2006
2
I have gone thru all my examples to find a solution to this problem... but no matter what I do I seem to get it wrong... I need help on how to simplify the following:

(xy + wz)(wx + yz)

This is what I have:
= xy'wx' + xy'yz' + w'zwx' + w'zyz' (Distributive)
= x' + y + w' + xx' + y + y' + zw + z' + w' + xw + z' + y' + z (DeMorgan's)
= .... (I keep simplifying)
= 1 (I know this isn't right...)

#### Papabravo

Joined Feb 24, 2006
16,163
After applying FOIL(First, Outer, Inner, Last) you should note that each term evaluates to zero.
Rich (BB code):
xywx'  -> 0 because xx' = 0
xy'yz  -> 0 because y'y = 0
w'zwx' -> 0 because w'w = 0
w'zyz' -> 0 because zz' = 0
Right?

#### Dave

Joined Nov 17, 2003
6,970
I'm with Papabravo on this one, from your line of analysis for the Distributive property you will find that the relations Papabravo has stated are present. Since you can't have something AND not-something then the expression is impossible (or in Boolean speak "false"/"0").

Dave

#### wilmania

Joined Nov 7, 2006
2
Fair enough... Thank you guys!

#### Papabravo

Joined Feb 24, 2006
16,163