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Every Boolean law has a dual?

  1. Apr 6, 2007 #1

    dfx

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    1. The problem statement, all variables and given/known data

    An extract from my notes reads that "every boolean law has a dual: any valid statement is also valid with:"

    . replaced with +
    + replaced with .
    0 replaced with 1 and vice versa.

    2. Relevant equations

    None

    3. The attempt at a solution

    I have no clue what this means. Surely it doesn't mean that you can switch around the + and . in any boolean statement and it will still hold true? Can anyone explain, and what are the implications of it?
     
  2. jcsd
  3. Apr 13, 2007 #2
    Sure it does. Suppose a + b = 1. The dual statement is a . b = 0 which is true. (Can you tell why?).
     
  4. Apr 13, 2007 #3

    Dick

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    It's not true at all. The dual statement to a+b=1 is (not a).(not b)=0. Replacing 0 by 1and vice versa implies you should replace a by (not a).
     
    Last edited: Apr 13, 2007
  5. Apr 14, 2007 #4
    Right. Sorry about that.
     
  6. Apr 14, 2007 #5
    Easiest way to analyze this is like this.

    a or b = 1

    not (a or b) =0

    now you break the bar(not) by changing the 'or' to an 'and' and you are left with:

    not a 'and' not b = 0

    This strategy of bar and breaking the bar is key in simplifying boolean equations. For much more complex eqns there is always the k-map.
     
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