# Simplify Boolean Algebra

Discussion in 'Homework Help' started by azall122, Feb 8, 2008.

1. ### azall122 Thread Starter New Member

Feb 8, 2008
1
0
(a+b'=c')(a+b'c)=a=b'c

(a'+b')(a'=b)=a'

(a'+a')'=a

2. ### luck New Member

Jan 25, 2008
8
0
You Will want to clarify what you are asking in the future. (Also proof read your posts before you make them; it took me a few minutes to figure out some of the = signs were suppose to be + signs)

(a+b'+c')(a+b'c)=a(a+b'+c')+b'c(a+b'+c')=a=b'c
First I factor (a+b'+c') into a and b'c.
From the first term we know if a is high then the expression will be high because we have a OR stuff.
We also know that if b'c is high then b'c*b' will also be high and because it is OR'ed with stuff the expression will be high

(a'+b')(a'+b)=a'(a'+b')+b(a'+b')=a'+a'b'+a'b+bb'=a'
First I factor (a'+b') into a' and b.
From the 4 terms I have we can factor an a' out to get a'(1+b'+b+0). Because of that 1 we know if a is low then the function will be high because we are OR'ing a' with stuff

(a'+a')'=(a*a)=a
This is straight up demorgan's. http://en.wikipedia.org/wiki/De_Morgan's_laws