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How to simplify this boolean algebra (logic)

  1. Oct 29, 2012 #1
    This is related to logic circuits.
    How do I simplify this boolean algebra(logic) into its simplest expression?

    h = (abc + abd + acd + bcd)’

    My first attempt would be to change it into product-of-sum.
    h = (abc)'(abd)'(acd)'(bcd)'

    and then what next?
     
  2. jcsd
  3. Oct 29, 2012 #2
    h = (abc)'(abd)'(acd)'(bcd)'

    = (a'+b'+c')(a'+b'+d')(a'+c'+d')(b'+c'+d')

    = (a'+b'+c'+d)(a'+b'+c'+d')(a'+b'+c+d')(a'+b'+c'+d')(a'+b+c'+d')(a'+b'+c'+d')(a+b'+c'+d') (a'+b'+c'+d')

    =(a'+b'+c'+d)(a'+b'+c+d')(a'+b+c'+d')(a+b'+c'+d')(a'+b'+c'+d')


    I hope it is correct..
     
  4. Oct 29, 2012 #3
    Could you explain to me how you add the new terms into the expression?
     
  5. Oct 29, 2012 #4
    Let's assume we have :

    F = A'+AB

    = A'(B+B') + AB # B+B'= 1, will not affect the equation.

    = A'B + (A'B' + AB) # Cancel each other.

    = A'B

    By the same way :)
     
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