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Boolean-Could someone check my answer?

  1. Feb 4, 2005 #1
    I was just wondering if someone could check my answer for simplifiying the below question.

    [itex]\bar{(A+B)}[/itex](A+B)= X the first (A+B) has a continuous bar above it from bracket to braket.


    Would this be the correct way to simplifiy it?
    ([itex]\bar{A}[/itex]*[itex]\bar{B}[/itex]) + (A+B)= X

    [itex]\bar{A}[/itex]A[itex]\bar{B}[/itex]+A[itex]\bar{B}[/itex]B= X

    (0)[itex]\bar{B}[/itex]+A(0)= X

    A[itex]\bar{B}[/itex]= X

    be the correct steps to simplifiy the problem?
     
    Last edited: Feb 4, 2005
  2. jcsd
  3. Feb 5, 2005 #2
    If only the first bracket has the bar over it then there seems to be a mistake in the first step. The sum can be converted into the product using deMorgan's Laws but you can do so only for the expression under the bar. So,

    [tex]\vec{A+B} = \vec{A}\vec{B}[/tex]

    Now do this all over and you should be through (the brute force method after the first step in most problems isn't a bad idea unless you observe some symmetry or vanishing terms...)
     
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