GATE Overflow | Digital Logic | Test 1 | Question: 29

Question

The minimum number of gates required to implement the Boolean function $(AB+C)$ if we have to use only 2-input NOR gate?

Answer 1

3 (A+B)(A+C)

Answer 2

We can write the finle result as

((A'+C')' + (B'+C')')' = F

Which is equal to 3 Nor gate required.