Boolean Logic Simplification Issue

Question

I hate this stuff. Just to note. + means OR * means AND ! means NOT.

(A+B) * (A+C) * (!B + !C)

(A | B) & (A | C) & (!B | !C) // more conventnal

The answer is A(!B + !C)

I'm trying to get there.

So I start off with using Distributive rule which gets me here (A + B) * C * (!B + !C)

and that's where I'm stuck. I know I some how have to get rid of B and C but I see no way using any of the rules. I've got Identity, Null, Itempotent, Inverse, Commutative, Associative, Distributive, De Morgan's, and Cancellation to work with.

Am I starting off wrong? I really just used the only rule that I could see possible to even use. I was horrible with doing Proofs in Geometry and this stuff just makes me feel like that all over again.

score 1 · Accepted Answer · answered Sep 24 '10 at 01:55

1

Your first step is wrong.

(A+B) * (A+C) is (A+(B*C)).

Next, (!B + !C) is !(B*C).

So we get A*(!(B*C)) + (B*C)*(!(B*C)), which gives the desired result.

answered Sep 24 '10 at 01:55

UncleO

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How do you go from (A+(B*C)) * !(B*C) to A*(!(B*C)) + (B*C) * (!(B*C)) I missed something big there. – Doug Sep 24 '10 at 02:05
1

@Doug - he's distributing (X + Y) * !Y to (X * !Y) + (Y * !Y). – dash-tom-bang Sep 24 '10 at 02:08
dash-tom-bang is right. Distribute the !(B*C) over the +. The point of that is we recognize (B*C)*(!(B*C)) is "false", so it drops out of the or statement. Convert !(B*C) back to (!B + !C) and you're done. – UncleO Sep 24 '10 at 02:16

score 1 · Answer 2 · answered Sep 24 '10 at 02:18

(A | B) & (A | C) & (!B | !C) = (A | (B & C)) & (!B | !C)
                              = (A | (B & C)) & !(B & C)

substitute D = (B & C)

                              = (A | D) & !D 
                              = A & !D
                              = A & !(B & C)
                              = A & (!B | !C)

Boolean Logic Simplification Issue

2 Answers2