# Boolean Algrebra help

Discussion in 'Homework Help' started by atlastx, Jul 1, 2012.

1. ### atlastx Thread Starter New Member

Jul 1, 2012
3
0
Hi All,

I'm new to this subject. I understand the basic laws and that the values of the variables can be either 1 or 0. Here is the formula I have, and the simplification I've done so far.

AB' + A( B' + C' ) + B( B' + C' )
= AB' + A( B' + C' ) + BB' + BC' <-- BB' = 0, removed
= AB' + AB' + AC' + BC' <-- AB' + AB' = AB'
= AB' + AC' + BC' <-- AB' + AC' = A(B' + C' )
= A(B' + C' ) + BC'

I have a feeling I have my groupings wrong. My problem with this so far is I'm not sure which laws to apply when. It all seems pretty haphazard.

Thanks!

2. ### atlastx Thread Starter New Member

Jul 1, 2012
3
0
I figured it out on my own... that was not fun... lol.

3. ### WBahn Moderator

Mar 31, 2012
17,715
4,788
The problem with "simplifying" a Boolean expression is that there are several competing and contradictory metrics by which "simple" can be measured. Fewest operators, fewest transistors, shortest delay, fewest gates with more than two inputs., etc. etc.

What you have done looks fine. You could regroup things a bit and, arguably, come up with something "simpler" by doing:

= AB' + (A + B)C'

One of the nice things about Boolean algebra is that it is pretty easy to check. Just make a truth table for the original expression and then an independent one for your final expression. If they are identical, then the expressions are equivalent, by definition. Whether it is "simplest" is a different issue.

I imagine it was the same way when you learned normal algebra (or just about any other technical subject). Only with experience and more than a little frustration will you start to internalize how to apply what rules to achieve what results -- and you'll very likely never be completely comfortably at it and will always have to "play with it" to some extent.

4. ### atlastx Thread Starter New Member

Jul 1, 2012
3
0
Thank you for the response. My final answer was B' + C'A, and my professor told me I was right.

5. ### WBahn Moderator

Mar 31, 2012
17,715
4,788
Always ask yourself if the answer makes sense. Perform as many sanity checks as you can.

In your final answer, you are saying that the result is True anytime that B is False, regardless of what A and C are.

Now look at the original expression:

Y = AB' + A( B' + C' ) + B( B' + C' )

If B is False, the third term is False and the first and second terms require A to be True. So these are NOT equivalent.

What about when B is True and C is False? The third term results in a True, but your final answer is False. Again, they are NOT equivalent.

Do the truth table.

Y = AB' + A( B' + C' ) + B( B' + C' )
Z = B' + C'A

 A B C Y Z 0 0 0 F T 0 0 1 F T 0 1 0 T F 0 1 1 F F 1 0 0 T T 1 0 1 T T 1 1 0 T T 1 1 1 F F