Multiplication gives
us distribution over the products, so
(a′+b+d′) (a′+b+c′+f′)
= a′ (a′+b+c′+f′) + b (a′+b+c′+f′) + d′ (a′+b+c′+f′)
And then you can
then distribute again each of the factors on the right.
Then you should simplify
in any given number of ways. To take as an example, you have a′b and ba′,
and since a′b + a′b = a′b + a′b = a′b, you can just drop one of them.
Since bb = b, you can rewrite bb as b and etc.
So in the end
part we should arrive at a sum of products. Then you can just invert. For
example, if at the end you had:
p′ = a′b + bc′ +
d′f ′+ a′f′
Then we would
have
p = p′′ = (a′b +
bc′ + d′f′ + a′f′)′ = (a′b)′⋅(bc′)′⋅(d′f′)′⋅(a′f′)′
Then applying De
Morgan's laws to each of the factors, e.g., (a′b)′ = a+b′, so we would
have
p = (a+b′)⋅(b′+c)⋅(d+f)⋅(a+f)
which is a
product of sums.
