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DeMorgan's theorem


Martin
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Now I know this may be lost on some off you, but I also know (hope) others understand it (because I don't)

 

Now a little key

 

 

. = AND

+ = OR

BOLD = NOT e.g D = NOT D

 

Original expression

(A.B) + (A.C) + (A.D)

 

DeMorganised expression

_________________

(A+B) . (A+C) . (A+D)

 

If anyone can tell me whether this is correct or not I'd me very appreciative! I'll even give you an interweb high five!

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Yep that is 100% correct. The way i was always taught at uni was that if you break the bar, you change the logic type (AND <-> OR gate). By removing the bar above a NOT gate, you also change it to NOT NOT (ie, the logic 1 or 0 doesnt change).

 

You can check your answer using a Karnaugh map as well if you want, but i wont go into that, youll have to google it!

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Thanks very much! I got the original expression from my karnaugh map and had to do DeMorgans on it.

 

It's hard because the woman teaching it isn't sure if she's doing it right so she'll tell you one way, next lessons she'll tell you it was wrong so teaches you her new way then next lesson she says go back to method1 :nutter:

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Thanks very much! I got the original expression from my karnaugh map and had to do DeMorgans on it.

 

It's hard because the woman teaching it isn't sure if she's doing it right so she'll tell you one way, next lessons she'll tell you it was wrong so teaches you her new way then next lesson she says go back to method1 :nutter:

 

And how much are you paying for this course?

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