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Boolean Algebra and DeMorgan's Theorems

Date: 05/14/2001 at 22:32:03
From: Craig Yates
Subject: Boolean algebra


Date: 05/15/2001 at 17:34:04
From: Doctor TWE
Subject: Re: Boolean algebra

Hi Craig - thanks for writing to Dr. Math.

I take it you want to simplify the expression by "breaking up" the 
inverters over the AB and the entire expression. The key is to use 
DeMorgan's theorems (sometimes called DeMorgan's laws), which state:

     ___   _ _     ___   _ _
     P.Q = P+Q     P+Q = P.Q

(Note that there's one inverter over the entire expression on the 
left-hand side of each equality, and separated inverters over each 
variable on the right-hand sides. Note also that I'm using '.' to 
represent AND, to avoid confusion)

The other key is to work from the outside in (or top down). First 
break up the inverter over the entire expression by treating the 
'not[A.(not B)]' as a single entity, call it X. Then we have:

which is equivalent to:
     _ _

Now substituting 'not[A.(not B)]' back in for X we get:

     ___   _   (we just "broke this up")
     A.B . C

and we can cancel the double-inverters over the 'A.(not B)' because 
they cover the exact same thing. (This is called double negation.) 
Then we have left:
       _   _
     A.B . C

I hope this helps. If you have any more questions, write back.

- Doctor TWE, The Math Forum   
Associated Topics:
High School Logic

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