Digital Logic - Implementing Boolean expression using minimum number of 2-input NOR gates

Question

Implement this Boolean expression using a minimum number of 2-input NOR gates. Then, illustrate with a clearly labelled logic circuit diagram.

F(w,x,y) = (x+y)(w+y)(x'+y')
         = [(x+y)' + (w+y)' + (x'+y')']' //double negation
         = [y'(x'+ w') + xy]'
         = [y'(xw)' + xy]'
         = [(y+xw)' + (x'+y')']'
         = [(y+(x'+w')')' + (x'+y')']'

As far as I know, NOR gate is (x+y)'. From here, I'm confused on how to begin using the result above to produce the logic circuit diagram.

Answer 1

Let's use the notation P ⊥ Q for nor, i.e., for (P + Q)'.

Start here:

1)

[(y+(x'+w')')' + (x'+y')']' = A ⊥ B

where A = (y+(x'+w')')' and B = (x'+y')'

2)

A = y ⊥ D

where D = (x'+w')' = x' ⊥ w'.

3)

B = x' ⊥ y'

4)

x' = x ⊥ 0
y' = y ⊥ 0

Now put everything together:

[(y+(x'+w')')' + (x'+y')']' = A ⊥ B
                            = (y ⊥ D) ⊥ (x' ⊥ y')
                            = (y ⊥ (x' ⊥ w')) ⊥ (x' ⊥ y')
                            = (y ⊥ ((x ⊥ 0) ⊥ (w ⊥ 0))) ⊥ ((x ⊥ 0) ⊥ (y ⊥ 0))

And I bet you will now be able to draw the circuit.