I need to express $(\bar A + \bar B)(A + B)$ using only 4 2-input NAND gates.
The hint given is: distribute the content in the second parentheses and then do De' Morgan's Law three times)
All I can think that means is: $$(\bar A + \bar B)(A + B)(A + B)$$
Applying De' Morgan's: $$(\overline{AB}) \overline{(\bar A \bar B)} \overline{(\bar A \bar B)}$$
There should be three top bars but I can't figure out how not to combine the last two… Is it the same thing regardless?
Is this even correct? Where can I go from here?
Best Answer
I figured it out just in case it helps anyone else. I was just dumb...
Go from: $(\bar A + \bar B)(A + B)$
To: $A(\bar A + \bar B) + B(\bar A + \bar B)$
Use De' Morgans a couple times and you're good!