Simplify $ABC+A'BC+AB'C+A'B'C'$ to logic gates

Question

I've simplified $ABC+A'BC+AB'C+A'B'C'$ to $BC+AC+A'B'C'$. However, I want to go further to logic gates for which ICs are readily available. I would like to use at most three such ICs.

Answer 1

The logic gates commonly implemented in TTL/CMOS ICs are two-input OR, AND and XOR, their negations and NOT. I will interpret "IC" in the question as one of these logic gates, so we are looking to use at most three logic gates to represent the given expression. Let's simplify it on paper first: $$ABC+A'BC+AB'C+A'B'C'$$ $$=(AB+A'B+AB')C+(A+B)'C'$$ $$=(A+B)C+(A+B)'C'$$ $$=(A+B)\odot C$$ where $\odot$ is the XNOR gate, returning 1 iff both inputs are equal.

Immediately we have the admissible representations (A OR B) XNOR C and (if XNOR is not available as a single chip) NOT((A OR B) XOR C).

Answer 2

Sorry this won't be strictly a math answer, but since you asked for IC circuits, then I would suggest you use Logic Friday, enter an expression, select the gates you want and it draws the minimal circuit.

For instance here with standard gates (thus excluding XNOR) you'll use $2$ of them :

(A nor B) xor C 

If you want only NAND and OR gates then you'll use $4$ of them :

((A or B) or C) nand ((A or B) nand C)

Note also that since XNOR is the mathematical equivalent of "$?\atop\iff$"

we can also use a $3$-input MUX gate which select the ouput according to its third entry:

MUX(u,v,s) = if (s=0) then u else v

(A or B) xnor C   is equivalent to   MUX( not(A or B), (A or B), C )