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Simplifying boolean expressions

  1. Sep 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Simplify the following Boolean expressions to a minimum number of literals
    (a+b+c')(a'b'+c)

    2. The attempt at a solution
    Whenever I tried this I made no progress in reducing the number of literals, I just reordered the expression.

    (a+b+c')(a'b'+c)
    =aa'b'+a'bb'+a'b'c'+ac+bc+cc'
    =0b'+a'0+a'b'c'+ac+bc+0
    =0+0+a'+b'+c'+ac+bc+0
    =a'b'c'+ac+bc
    =a'b'c'+c(a+b)

    I began with 6 literals and ended with 6. What else can I try?
     
  2. jcsd
  3. Sep 17, 2012 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    Are you allowed to use XOR?
     
  4. Sep 17, 2012 #3
    It doesn't say we can't and we did cover it, however it doesn't show up in any of the other problems or examples so far
     
  5. Sep 17, 2012 #4
    I tried to use X(N)OR and this is what I came up with:

    From
    =a'b'c'+ac+bc

    I can multiply ac and bc by (1) in the form (x+x')

    =a'b'c'+ac(b+b')+bc(a+a')
    =a'b'c'+abc+ab'c+abc+a'bc

    Get rid of one of the two abc (redundant)

    =a'b'c'+abc+ab'c+a'bc

    Factor

    =bc(a+a')+b'(ac+a'c')
    =bc(1)+b'(aXORc)'
    =bc+b'(aXORc)'

    Now I have 5 literals, but is that really the most simplified term? I did all this work just to eliminate 1 literal :(
     
  6. Sep 18, 2012 #5

    NascentOxygen

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    Staff: Mentor

    = c'·a'b' + c(a+b)

    = c'·a'b' + c·(a'b')'

    = ....
     
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