Understanding Demorgan's Theorem for Logic Gates

[hell18]

when using Demorgan's Theorem i didn't understand why it had to be used as logic gates AND or OR?.

e.g.

not x and y not x or not y


in other words when facing a situation with 2 logic gates the same you make them the opposite of each other to get the answer?

 

[StephenPrivitera]

Your post is difficult to understand. DeMorgans shows how the negation distributes over the conjunction and disjunction.
For example
~(A & B) <--> (~A V ~B)
This makes sense. The left hand side says
"It is not the case that both A and B are true."
The right hand side says,
"Either A is not true or B is not true."
Clearly these say the same thing since if A and B aren't both true, then one of them has to be false. And if one of either A or B is false, then they can't both be true.

 

[Legolas]

Well I'm not quite sure what you are asking either, but maybe I can help.

when using Demorgan's Theorem i didn't understand why it had to be used as logic gates AND or OR?.

The only answer I can come up with would be to restate the definition of these gates, and/or restate how they operate. I don't really see the point of this so I will move on.


in other words when facing a situation with 2 logic gates the same you make them the opposite of each other to get the answer?

Well, I'm not sure what you mean by "the answer", but by applying DeMorgan's theorem you will have restructured your logic to meet whatever requirements were initially set out. If you are using a technique called "Technology Mapping" (which in most cases is used to convert all your logic to NAND or NOR gates) then DeMorgan's is the theorem you would apply. Usually these gates are faster and hence the reason you might be looking for different logic.

 

[PPM48]

Funny, I was just thinking of this on the way home and then used GOOGLE and found this

site.

Some of the folks, on the august site, were having some problems with this, so I thought it

through and here is what I concluded. Anything I did incorrectly, please advise!

Proving DeMorgan's Law:

To Prove: (AB)'=A'+B'

From AB, then (AB)' will be TRUE for all other combination of AB

i.e.: A'B+AB'+A'B' are TRUE.

A'B+AB'+A'B'= A'B+AB'+A'B'+A'B'=B'(A+A')+A'(B+B')=A'+B'

General Expansion Case:

(ABC)'=A'+D' [SETTING BC=D]=A'+(BC)'=A'+B'+C'