DeMorgan's Theorem Question

No. You are basically saying that AND == OR. (A' B C)+(A B' C)+(A B C') is the equivalent of ((A'BC)'(AB'C)'(ABC')')'

DeMorgan's Theorem says that (AB)' = A' + B', or for three variables (XYZ)' = X'+Y'+Z'

If you substitute: X=(A' B C) Y=(A B' C) Z=(A B C')

Using truth tables:

F1 = (A' B C) + (A B' C) + (A B C') F2' = (A'BC)'(AB'C)'(ABC')'

A B C | X Y Z | F1

-----------+-----------+---- 0 0 0 | 0 0 0 | 0 0 0 1 | 0 0 0 | 0 0 1 0 | 0 0 0 | 0 0 1 1 | 1 0 0 | 1 1 0 0 | 0 0 0 | 0 1 0 1 | 0 1 0 | 1 1 1 0 | 0 0 1 | 1 1 1 1 | 0 0 0 | 0

A B C | X' Y' Z' | F2'| F2

-----------+-----------+----+---- 0 0 0 | 1 1 1 | 1 | 0 0 0 1 | 1 1 1 | 1 | 0 0 1 0 | 1 1 1 | 1 | 0 0 1 1 | 0 1 1 | 0 | 1 1 0 0 | 1 1 1 | 1 | 0 1 0 1 | 1 0 1 | 0 | 1 1 1 0 | 1 1 0 | 0 | 1 1 1 1 | 1 1 1 | 1 | 0

F1 = F2 Q.E.D.

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No, inside that expression is A' A = 0. Do you mean (A' + B + C)(A + B' + C)(A + B + C')

Then from that use distributive law of * over + to see what you get.

X+Y+Z = NOT(NOT(X).NOT(Y).NOT(Z))