F = (A + B)(B'C)(A' + C')
I have to convert this expression using only NOR gates. I am stuck on the format of this expression since it is not in SoP form.
I tried to distribute (B'C) into the expression and I got:
(B'C)+(B'A)+(CA')
Then I took DeMorgan's Law of the expression and i got:
(B+C')(B+A')(C'+A)
My Question is, can I now take this expression and apply double negate it so I can use NOR gates? Is this correct?
$$ \overline{\overline{(B+C')}+\overline{(B+A')}+\overline{(C'+A)}} $$
Yes, that's right:
(B+C')(B+A')(C'+A) = (Double negation)
[(B+C')(B+A')(C'+A)]'' = (DeMorgan)
[(B+C')'+(B+A')'+(C'+A)']'
... which is exactly your
$$ \overline{\overline{(B+C')}+\overline{(B+A')}+\overline{(C'+A)}} $$
Also, if you can't work with C', and have to work with C (that is, if you can't use Not's), use the fact that:
C' = C NOR C
... Unfortunately what you did at the beginning is not right ...
(A + B)(B'C)(A' + C') does not work out to
(B'C)+(B'A)+(CA')
and you certainly can't just negate that to do a DeMorgan on that to get:
(B+C')(B+A')(C'+A)
since you'd just be adding a negation to the whole expression out of nowhere!
Instead:
(A + B)(B'C)(A' + C') = (Association)
(A + B)B'C(A' + C') = Reduction x 2
AB'CA' = False!
And an easy NOR formula for that is: NOR(A,A') = NOR(A,NOR(A,A))
I wonder though ... should that initial expression maybe be:
F = (A + B)(B'+C)(A' + C') ?
Because then you can do what I did above:
(A + B)(B'+C)(A' + C') =
[(A + B)(B'+C)(A' + C')]'' =
[(A + B)'+(B'+C)'+(A' + C')']' =
NOR(NOR(A,B'),NOR(B',C),NOR(A',C')) = (if needed)
NOR(NOR(A,NOR(B,B)),NOR(NOR(B,B),C),NOR(NOR(A,A),NOR(C,C)))