Simplifying Simple Boolean XOR Expression (!AB + A!B)

Question

I am trying to simplify the 5 gate XOR from a A!B + !AB expression to a (A + B)!(A + B) implementation. How can I convert expressions like these requiring 5 gates to a simpler 3 gate implementation when the expressions are equivalent? What are the mathematical steps?

Thanks

Answer 1

If I understood your question correctly do this...

Use DeMorgan's rule to rewrite:

$A!B = !(!A+B)$

Then write it together

$A!B + !AB = !(!A+B) + !(A+!B) $

use DeMorgan again

$!(!A+B) + !(A+!B) = !(!A+B)(A+!B)$