# Simplifying Boolean Expression

Discussion in 'Homework Help' started by ramla, Dec 22, 2008.

1. ### ramla Thread Starter New Member

Dec 22, 2008
2
0
Hi,

I have an expression of nine literals as follows:

BC+AC'+AB+BCD.

I have to reduce it to four literals.

This is the way I proceeded.

BC+AC'+AB+BCD = BC(1+D)+AC'+AB
= BC+AC'+AB

I have reduced it to six literals. But I am not able to proceed further.

Thanks and Regards,
Ramla

Dec 18, 2008
11
0
It cannot be simplified more than that, I think. But wait for a confirmation from an expert

3. ### vvkannan Active Member

Aug 9, 2008
138
11
I think using K-map we will get a more simplified answer.

4. ### Ratch New Member

Mar 20, 2007
1,068
3
ramla,

vvkannan,

Correct, a K-map or one of the tabulation methods is the way to go. Doing it with Boolean algebra is the pits. Once you have done it with the K-map or any other closed method, you will easily be able see what terms make up the final solution, and then back compute your Boolean algebraic method to get the minimal solution.

Behold:

BC+AC'+AB
BC+AC'+AB(C+C')
BC+AC'+ABC+ABC'
BC(1+A)+AC'(1+B)
BC+AC'

Ratch

Last edited: Dec 22, 2008
5. ### ramla Thread Starter New Member

Dec 22, 2008
2
0
Thanks a lot ratch..