DeMorgan's Theorem Question

Can (A' B C) + (A B' C) + (A B C') be converted to (A' B C)(A B' C)(A B C') using DeMorgans Theorem. I think not but I have been told that this can be done. If it can be done can anyone explain how to do this. IMO even the truth tables would be different. TIA

Reply to
BIGEYE
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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.

Reply to
krw

I wanna talk about my dick.

Reply to
Mincer

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.

Reply to
William Elliot

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

Reply to
Airy R. Bean

Reply to
BIGEYE

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