# Solving Boolean Expressions with Theorems

I'm having the hardest time wrapping my head around this stuff. This is a homework problem, one of many. I just need some help on what to do.

BC + A'B' + A'C' = ABC + A'


I've tried this:

BC + A'(B' + C') = BC + A


After this my erase marks get more frequent. It seems to me none of the theorems provided offer obvious plans of attack.

SOLVED:

I think the idea is to identify was to get at least one thing matching on either side and work from there with some guess and check. With help, I found:

BC + A'B' + A'C' = ABC + A'
BC(A' + A) + A'B'(C' + C) + A'C'(B' + B) = ABC + A'
ABC + A'BC + A'B'C' + A'B'C + A'B'C' + A'BC' = ABC + A'
ABC + A'BC + A'B'C' + A'B'C + A'BC' = ABC + A'
ABC + A'(BC + B'C' + B'C + BC') = ABC + A'
ABC + A'(B(C + C') + B'(C' + C)) = ABC + A'
ABC + A'(B + B') = ABC + A'
ABC + A' = ABC + A'

• There are only three variables here. Have you tried making a Karnaugh map?
– MJD
Sep 18, 2013 at 3:39
• I must prove it with theorems. I wish I could use K-Maps or truth tables, they are much easier to see. Sep 18, 2013 at 3:41
• What does "with theorems" mean? What theorems are you allowed to use? Sep 18, 2013 at 3:43
• Presumably standard stuff like commutativity and associativity of $\lor$ and $\land$, distributivity, de Morgan's laws, absorption laws, and the like.
– MJD
Sep 18, 2013 at 3:45

Hint: On the left side, rewrite $BC$ as $(A' + A)BC$.
• Yes, and the right-hand side as $ABC$, so you need to get $ABC$ in there somehow, and this will do it.