Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having a very hard time simplifying this:

A!B!C+ABC+!A!BC+!AB!C

The objective here is to simplify the equation until it can be expressed in "AND"'s "OR"'s and "NOT"'s. I have to create an integrated circuit for the simplified expression using only those 3 gates in Logisim.

This is as far as I've gone:

A(!B!C+BC)+!A!BC+!AB!C

A+!A!BC+!AB!C

A+!BC+B!C

I'm not sure if this is right

share|improve this question
3  
It might help if you said what $!A$ means. At least I don't know, and at first thought it might mean "not", but the paragraph afterward suggests it isn't "not". Is it mayby xor or something binary which isn't typically used in math, but only in circuit design? –  coffeemath Nov 10 '12 at 4:31

1 Answer 1

If $!X$ means $1+X$ in the Boolean algebra then your expression simplifies to $A+B+C$. Check this by replacing the $!$ operators as indicated and distributing the multiplications. This expression can be written as

$$ (A \operatorname{AND} \operatorname{NOT} (B \operatorname{OR} C)) \operatorname{OR}\\ (B \operatorname{AND} \operatorname{NOT} (A \operatorname{OR} C)) \operatorname{OR}\\ (C \operatorname{AND} \operatorname{NOT} (A \operatorname{OR} B)) \operatorname{OR}\\ (A \operatorname{AND} B \operatorname{AND} C). $$

Note that $X+Y$ can be written as $(X \operatorname{OR} Y) \operatorname{AND} \operatorname{NOT} (X \operatorname{AND} Y)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.