Boolean Algebra and DeMorgan's TheoremsDate: 05/14/2001 at 22:32:03 From: Craig Yates Subject: Boolean algebra ____ __ _ AB+C 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: ___ X+C which is equivalent to: _ _ X.C 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 http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/