Simplify this using DeMorgan's Laws

How many gates?

1 AND

1 OR

3 NOT

steps:

break the line change the sign + becomes * break the line change the sign YZ becomes Y + Z remove double nots distribute X

Gates needed if distributed:

2 AND

1 OR

1 NOT

Gates needed if not distributed:

X(!Y+Z)

1 AND, 1 OR, 1 NOT