Implementing Boolean function, F using no more than two NOR gates

Question

[Steps provided by solution manual] Implement Boolean function, F using no more than two NOR gates.

1. F = AC' + A'D' + B'CD'
2. F' = D + ABC
3. F = [D + ABC]' = [D + (A' + B' + C']')]'

Hi. I had this as a homework problem and I don't understand how the function complemented from F = AC' + A'D' + B'CD' to D + ABC.

What I had is the following:

F'=(AC' + A'D' + B'CD')' = (AC')'(A'D')'(B'CD')' =(A'+C)(A+D)(B+C'+D)

I have no idea how to I move on to next step any further. Am I missing any boolean theorems?

Also, I would like to know why the solution manual complemented the function in the first place.

Answer 1

I'd ask you to check the problem again. A NOR gate has two inputs and one output. If all you have is two NOR gates, you can't build anything with an output that depends on more than three inputs. The most complex thing you could build is NOR (NOR (a, b), c).

I start to get it... Your textbook seems to be generous in what it calls NOR. I learnt that NOR (A, B) = (A + B)'. Your textbook seems to accept more than one input, [D + (A' + B' + C')']' can be done with two NOR's and complementing the input: Z = NOR (A', B', C'), F = NOR (D, Z). The step from 1 to 2 is just wrong.