Boolean Algebra help...

Thread Starter

avenger09123

Joined Feb 5, 2009
1
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??????
 

Ratch

Joined Mar 20, 2007
1,070
avenger09123,

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

F = A'BD' + AB'C + AB'D' which is your answer.

Ratch
 

Ratch

Joined Mar 20, 2007
1,070
photonxyz,

original problem

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

Simplifies to the following;

A'BD'+AB'D'+ABC
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
 
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