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Demorgan's Theorem

  1. Mar 5, 2008 #1
    [tex]\overline{ \overline{A}+ \overline{B} + \overline{A}B }[/tex]

    [tex]\overline{ \overline{A}+ \overline{B} } * \overline{ \overline{A}B }[/tex]

    [tex]\overline{ \overline{A}}* \overline{\overline{B} } * (\overline{ \overline{A}}+\overline{\overline{B }})[/tex]

    [tex]AB(A + \overline{B}) [/tex]

    [tex]AAB + AB\overline{B} [/tex]

    ANSWER=AB

    Just wanted to check. I haven't done this in a while
     
  2. jcsd
  3. Mar 5, 2008 #2
    The third step is wrong, but you got the right answer.
     
  4. Mar 6, 2008 #3
    I see the mistake. It should be:

    [tex]\overline{ \overline{A}}* \overline{\overline{B}} * (\overline{ \overline{A}}+\overline{B})[/tex]

    [tex]AB(A + \overline{B}) [/tex]

    [tex]AAB + AB\overline{B} [/tex]

    [tex](A)B + A(0) [/tex]

    [tex]AB[/tex]

    Thanks for pointing that out
     
  5. Mar 7, 2008 #4
    I put this equation in a K-map and I was unable to simplify it. Is there anyway to do this with exclusive or? Thanks for the help

    \overline{A1}\overline{A0}\overline{B1}\overline{B0} + \overline{A1}A0\overline{B1}B0 +
    A1\overline{A0}B1\overline{B0} + A1A0B1B0
     
  6. Mar 7, 2008 #5
    Oops the equation should go as follows:

    [tex](\overline{A1}*\overline{A0}*\overline{B1}*\overline{B 0}) + \overline{A1}A0\overline{B1}B0 + A1\overline{A0}B1\overline{B0} + A1A0B1B0 [/tex]
     
  7. Mar 10, 2008 #6
    Any suggestions?
     
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