# Boolean Algebra

Hi there,

I've never done this type of Algebra... and it's a bit confusing. I've completed this question, but I don't know if I took the right steps...

*=AND, +=OR, '=NOT

If A*B = 0, A+B=1

Prove: (A+C)*(A'+B)*(B+C) = B*C

I expanded it a bit and applied some of the rules. Can someone explain it step by step? I got the answer but I'm a bit confused at how I got the answer.

Thanks!

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AKG
Homework Helper
(A+C)*(A'+B)*(B+C)
= (A+C)*(B+C)*(A'+B) (commutation)
= ((A*B)+C)*(A'+B) (distribution)
= (0+C)*(A'+B) (substitution A*B = 0)
= C*(A'+B)

Now I don't know what "rules" you know or how formal you have to be, but you should be able to see that (A'+B) = B given A+B = 1. Clearly, if A'+B = 0, then B = 0. If A'+B = 1, we want B to also necessarily be 1. But suppose it's not, so suppose A'+B=1 but B=0. Then A' = 1, which means A=0. So both A and B would be 0, but we're given A+B = 1, which is a contradiction, so B does indeed have to be 1 if (A'+B) is 1.

Mhm.. I understood all the steps until the last one...this is what I think...

Since AB = 0 and A+B=1

That leaves us with two choices for A or B. A can either be 0 or 1, and B must be the exact compliment.

Now we're stuck with C(A'+B). *Here's where I'm a bit confused*

If A is 0, B will be 1, then A' + B = 1
If A is 1, B will be 0, then A' + B = 0

From which of these two can we say C*(A'+B) = B*C?

in otherwords... how does (A'+B) = B?

Last edited:
AKG
Homework Helper
Well I've already told you how (A' + B) = B. However, if you didn't follow the explanation, then consider what you said yourself:

A can either be 0 or 1, and B must be the exact compliment.

So B = A', right? So (A' + B) = B + B = B (Idempotence).

Ah... I think I understand what I just said. A'=B.. then you'd just have to sub it in.

Man, boolean algebra is mighty weird! I just started a week ago, still getting the hang of it. Thanks for your help!

how about a proof of A'=B

given A*B=0
(or with A')
A'+A*B=A'
(and with B)
A'*B + A*B*B=A'*B
=> (A'+A)*B=A'*B
=> B=A'*B (1)

given A+B=1
(and with A') => A'*A + A'*B = A'
=> A'*B = A'

(substitute from (1) ) => B=A'