How to apply De Morgan's theorem to a boolean equation?

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The discussion focuses on applying De Morgan's theorem to a boolean equation derived from a circuit diagram. Participants clarify that the initial expression provided, F = abC + AbC + aBC, may not be suitable for demonstrating De Morgan's laws. They emphasize the importance of correctly identifying the boolean expression before applying the laws, which state that negating an AND operation results in an OR operation of the negated variables, and vice versa. Additionally, the conversation highlights that the expression can be simplified without using De Morgan's laws. Ultimately, the goal is to find a suitable formula that requires De Morgan's laws for simplification.
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Homework Statement



http://img525.imageshack.us/i/circuitg.png/

Get boolean equation and apply De Morgan.

Homework Equations


The Attempt at a Solution



I'm not sure how to do this. Treat CLK like an variable? If so, is this correct?

F = abC + AbC + ABC

I do not know. I suppsoe you then apply DeMorgan.
 
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I think you must get a formula like X1 = ~(A and C), then simplify it using De Morgan's laws and possibly factor out C at the end.
 
huh? I don't know what you're talking about, but that's a boolean expression. I first need to be sure I have the correct boolean expression.
 
What is boolean algebra? It is a set of rules that allow you to manipulate boolean expressions and equations. De Morgan's laws are two of the many rules in boolean algebra. I think this question is about using De Morgan's laws to simplify a boolean expression.

The expression you found is not the formula the author of the question wanted you to find. Your formula can be simplified using boolean algebra, but it does not require De Morgan's laws to simplify. So if you want to see De Morgan's laws in action, you need another formula.

F = abC + AbC + aBC

This can also be written:

F = (~A and ~B and C) or (A and ~B and C) or (~A and B and C)

where the ~ means "not" or "= 0". The benefit of writing it this way is that De Morgan's laws can be given:

1. ~(A and B) = ~A or ~B
2. ~(A or B) = ~A and ~B

You couldn't write either of these using capital and small letters.

But your formula does not need these laws to be simplified. If you look at X1 in the circuit, it is a combination of A and CLK according to that logic gate. One might say:

X1 = ac + aC + Ac

or one might say

X1 = ~(A and C)

These are just the same. If you can see that these are the same, then consider how I found this formula. It is just the definition of a NAND gate.

And now you should be able to use the logic gates in the picture to build up the formula that the author wanted you to find, the one that needs De Morgan's laws to simplify.
 

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