Discussion in 'Homework Help' started by wilmania, Nov 7, 2006.

1. ### wilmania Thread Starter New Member

Nov 7, 2006
2
0
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...)

2. ### Papabravo Expert

Feb 24, 2006
10,148
1,791
After applying FOIL(First, Outer, Inner, Last) you should note that each term evaluates to zero.
Code ( (Unknown Language)):
1.
2. xywx'  -> 0 because xx' = 0
3. xy'yz  -> 0 because y'y = 0
4. w'zwx' -> 0 because w'w = 0
5. w'zyz' -> 0 because zz' = 0
6.
Right?

3. ### Dave Retired Moderator

Nov 17, 2003
6,960
144
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

4. ### wilmania Thread Starter New Member

Nov 7, 2006
2
0
Fair enough... Thank you guys!

Feb 24, 2006
10,148
1,791