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Boolean Algebra Simplification

  1. Feb 21, 2015 #1
    ab'c + a'b + bc' + abc

    = ac + a'b + bc' (How to further reduce this?)

    Kmap gives B + AC
     
  2. jcsd
  3. Feb 21, 2015 #2

    phinds

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    Do you know about karnaugh maps? The reduction to B + AC is trivial and obvious if you do.
     
  4. Feb 21, 2015 #3
    As in is there a boolean algebra way? Yea i know it can be done in kmap. Just curious
     
  5. Feb 21, 2015 #4
    You can reduce ac + a'b + bc' further using DeMorgan's theorem as a first step.
     
  6. Feb 21, 2015 #5

    phinds

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    Sorry, yes, your subject line does say algebraic simplification. As milesyoung said, use deMorgan's theorems.
     
  7. Feb 21, 2015 #6

    phinds

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    One thing I should add: if you can reduce the complexity of a Boolean statement using a K-map then you can ALWAYS do the same thing algebraically. It would not make any sense for it to be otherwise. It may not be as easy as w/ a K-map but it has got to be doable.
     
  8. Feb 22, 2015 #7

    NascentOxygen

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    First question: do you know De Morgan's theorems?

    If you can reduce your logic expression to ac + b.¬(ac) I can give you the next step after that, if needed.
     
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