Q:

how does A'B'C' convert to A'+B'+C'?

I'm trying to learning boolean algebra.
One of de morgan's law is not understood.
how does A'B'C' convert to A'+B'+C'?
As I know,
1. (A'B')' = A+B
2. (AB)'=A'+B'
3. (A+B)'=A'B'
I think it should be (A+B+C)'.
Is this same between A'+B'+C' and (A+B+C)' or A'B'C' and (ABC)'? 
http://www.indiabix.com/digital-electronics/boolean-algebra-and-logic-simplification/discussion-138

A:

\$\bar A \bar B \bar C \$ does not equal  \$  \bar A+ \bar B + \bar C \$
If you let A=0, B= 1, and C=1
Then \$\bar A \bar B \bar C \$ = 0 and \$  \bar A+ \bar B + \bar C \$ = 1
It's clear that they are not the same
If you apply deMorgan's for \$\bar A \bar B \bar C \$ you would end up with  (A+B+C)'. Which is what you said.

A:

In your post, you say:
2. (AB)'=A'+B' 
3. (A+B)'=A'B'
I think it should be (A+B+C)'.  <== I don't know where this fits in

then you provide four combinations:
A'+B'+C' and (A+B+C)' or A'B'C' and (ABC)' 

Looking at each one,
A'+B'+C' = (ABC)'     (flipping 2 around and extending to three terms)
(A+B+C)' = A'B'C'     (from 3, extending to three terms)
A'B'C' = (A+B+C)'     (flipping 3 around and extending to three terms)
(ABC)' = A'+B'+C'     (from 2, extending to three terms)

so the first and last are the same, and the middle two are the same.

