# Thread: Trouble with de morgans law and boolean algebra

1. ## Trouble with de morgans law and boolean algebra

So Ive attached an example of a couple of questions from a past exam paper. But, I have no idea how the answers can be achieved. If anyone can take a look, and explain step by step how to achieve each answer, that would be great, as I've tried and still dont get it :-|

2. ## Re: Trouble with de morgans law and boolean algebra

Do you know what these things mean? What does "$\displaystyle \overline{A}$" mean? What does "A+ B" mean? What does "$\displaystyle A\cdot B$" mean?

3. ## Re: Trouble with de morgans law and boolean algebra

I know that " \overline{A}" means "not A" and "A+ B" means "A or B" and " A\cdot B" means "A and B" but I still have trouble with arriving at the answer

4. ## Re: Trouble with de morgans law and boolean algebra

use the following
for first part B.(A+COMP(A))
U KNOW A+COMP(A)=1
AND B.1=B
FOR SECOND PART TAKE OUT B AS COMMON
U GET B(1+A)
1+A=1 AND B.1=B
FOR THIRD PART COMP(X+Y)=COMP(X)COMP(Y)
HERE X=B AND Y=COMP(A)+COMP(B)
SO COMP(X)COMP(Y) IS GIVEN FIND COMP(X+Y)
COMP(X+Y)=COMP(B+COMP(A)+COMP(B))
NOW B+COMP(B)=1 AND 1+COMP(A)=1
ALSO COMP(1)=0