NOTE: $b'$ means $b$ not
I'm trying to convert $ab'd + ab'cf$ to product of sums form
My professor gave us the following hint:
"Invert the equation, reduce it to sum-of-products, then invert it again. The result will be the original equation, but in product-of-sums form."
Now, I think I have the inverting portion pretty down pat. I just use De Morgan's laws.
So $p' = (a'+ b + d') \cdot (a' + b + c' + f')$
However, I'm not comfortable with the next step.
Does $p'$ expand to: $$(a'a' + a'b + a'c' + a'f') + (ba' + bb + bc' + bf') + (d'a' + d'b + d'c' + d'f')\ ?$$
Ignoring the optimization for now...
Then $p' =\ldots $?
Do I do this? $p' = ( (a'a' + ba') + (a'a' + bb)\cdots$ and so on?
I feel like I'm doing this wrong.
Any help would be greatly appreciated. I'll respond quickly!