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Help with simplifying boolean expression

  1. Jan 29, 2015 #1
    < Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

    (A+B)&(C+D) + (A+B)&(C+D)' + C
    (A+B)&(C+D) + (A+B)&(C'&D') + C by deMorgans
    (A+B)&[(C+D)+(C'&D')] + C by Distributive

    I'm just wondering if I did anything wrong in this simplification or if it can be simplified any further.
     
    Last edited by a moderator: Jan 29, 2015
  2. jcsd
  3. Jan 29, 2015 #2

    Svein

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    Well, use the distributive law on the first part of first line - (A+B)&(C+D) + (A+B)&(C+D)' = (A+B)&((C+D) + (C+D)') = (A+B) (since P + P' = TRUE).
     
  4. Jan 29, 2015 #3
    Thanks, I didn't even see that.
     
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