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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'
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  • $\begingroup$ There are only three variables here. Have you tried making a Karnaugh map? $\endgroup$ – MJD Sep 18 '13 at 3:39
  • $\begingroup$ I must prove it with theorems. I wish I could use K-Maps or truth tables, they are much easier to see. $\endgroup$ – Jerrod Sep 18 '13 at 3:41
  • $\begingroup$ What does "with theorems" mean? What theorems are you allowed to use? $\endgroup$ – Alex Kruckman Sep 18 '13 at 3:43
  • $\begingroup$ Presumably standard stuff like commutativity and associativity of $\lor$ and $\land$, distributivity, de Morgan's laws, absorption laws, and the like. $\endgroup$ – MJD Sep 18 '13 at 3:45
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Hint: On the left side, rewrite $BC$ as $(A' + A)BC$.

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  • $\begingroup$ ohh, because A + A' is 1? $\endgroup$ – Jerrod Sep 18 '13 at 3:57
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    $\begingroup$ Yes, and the right-hand side as $ABC$, so you need to get $ABC$ in there somehow, and this will do it. $\endgroup$ – MJD Sep 18 '13 at 3:58

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