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Simplify boolean expression (a+b+c)*(a'+c)*(a'+b')

  1. Oct 9, 2011 #1

    dba

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    1. The problem statement, all variables and given/known data
    I need to simplify the boolean expression by algebraic manipulation as much as possible.
    (a+b+c)*(a'+c)*(a'+b')

    2. Relevant equations
    + stands for OR
    * stands for AND (or no operator between variables)
    ' Stands for NOT

    3. The attempt at a solution
    I tried but I do not know if this is correct.

    (a+b+c)*(a'+c)*(a'+b')
    = [aa' + ac + ba' + bc + ca' + cc] * (a'+b') => aa'=0, cc=c
    = [0 + ac + ba' + bc + ca' + c] * (a'+b') => c + bc = c
    = [ac + ba' + ca' + c] * (a'+b') => c + ac = c
    = [ba' + ca' + c] * (a'+b') => c + a'c = c not sure if that works
    = [ba' + c] * (a'+b')
    = ca' + cb' + a'a' + a'b' + ba' + bb' => a'a'=a', bb'=0
    = ca' + cb' + a' + a'b' + ba' => factor out the last two a'
    = ca' + cb' + a' + a'(b' + b) => b'+b=1
    = ca' + cb' + a' + a' => a'+a'=a'
    = ca' + cb' + a' => factor out a'
    = a'(1+c) + cb' => 1+c = 1
    = a+cb'

    Thanks for any help!
     
  2. jcsd
  3. Oct 10, 2011 #2

    NascentOxygen

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    Staff: Mentor

    I think you're right this far, but I can't follow your working thereafter.

    To check your own work, you can draw up a truth table. The expression you are given should always evaluate to the same values as your "simplified" version. If in an exam, you don't have time for a complete TT, try just a few values, e.g., a=b=1 c=0
    then original expression is 0; your expression evaluates to 1.
     
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