# Boolean Algebra help...

Discussion in 'Homework Help' started by avenger09123, Feb 5, 2009.

1. ### avenger09123 Thread Starter New Member

Feb 5, 2009
1
0
Ok here's the problem....
(A'+B'+C)(A+D')(A'+B+D')(A+B)(A+C+D')

And here's my work....
(A'+B'+C)(A+D')(A'+B+D')(A+B)(A+C+D')

[A'D+A(B+C)] [A'B+B(A+D')] (A+C+D')
-----Did the stuff inside the brackets now------

(A'D+AB+AC)(A'B+AB+D'B)(A+C+D')

(A'D+AB+AC)(B+A'BC+A'BD'+AB+ABC+ABD'+ABD'+BCD'+B)
^^^^^^^^eliminate uneccessary doubles
(A'D+AB+AC)(AB+A'BD'+BC+ABD'+B+BCD')

OK WHERE DO I GO FROM HERE? OR DID I SCREW UP AND WHERE??????

2. ### Ratch New Member

Mar 20, 2007
1,068
3
avenger09123,

I will let you decide that.

F = (A'+B'+C)(A+D')(A'+B+D')(A+B)(A+C+D')

F' = ABC' + A'D + AB'D + A'B' + A'C'D by DeMorgan's theorem

Plot the F' terms on a Karnaugh map with a "x" mark.

Read the 6 terms on the K-map which have not been marked. These are the simplified F terms.

Ratch

3. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Better to use a K-map to minimize such quite big problems as you will get the optimum solution.

4. ### photonxyz New Member

Feb 13, 2009
3
0
original problem

(A'+B'+C)(A+D')(A'+B+D')(A+B)(A+C+D')

Simplifies to the following;

A'BD'+AB'D'+ABC

5. ### Ratch New Member

Mar 20, 2007
1,068
3
photonxyz,

Yes, you are right. I should not have put a "'" on the B variable of the ABC term. I wonder why the OP did not catch that mistake.

Ratch