Boolean Algebra
posted by Anonymous .
I need help starting this, I don't want a plain answer just a what to do.
(bc'+a'd)(ab'+cd')Simplify
My best guess is to use the distributive
property.
(a+b)(a+c)= a(b+c). But now that there is a 4th variable what do I do?

use it twice
(bc'+a'd)(ab'+cd')
= bc'(ab'+cd') + a'd(ab'+cd')
= abb'c' + bcc'd' + aa'b'd + a'cdd'
Now, since xx' = 0, the result is 0 
OOK, thank you, I see how you did that step. that makes perfect sense.
so simplified you get
ac'+bd'+b'd+a'c 
no, the result is 0, since
abb'c = a(bb')c = a0c = 0
each term is 0