Two points: 1. Aquaint yourself with DeMorgan's Theorums. 2. Remember that you can apply the double-NOT function such that it will not change the expression, i.e. A => NOT gives A' => NOT gives A (where ' = NOT). This is very useful when applying DeMorgan's Theorums. Have a go, post up your attempts and we can help you from there. Dave
Did you read the information in the link? If you can cite the specific parts you are having trouble with, we can walk you through them.
Almost... you dropped an "A" Needs to be A(B'+C') instead of simply B'+C' Otherwise, you got it right!
DEAR you can use simple one law ( A . B )' = A' + B' A( B. C' +(D.E)' )+ (B.F)' = A( B. C' + D' + E' ) + B' + F' = A . B . C' + A .D' + A .E' + B' + F'