demorgans theorem: need more help

Discussion in 'Homework Help' started by mik3ca, Feb 13, 2007.

  1. mik3ca

    Thread Starter Active Member

    Feb 11, 2007
    189
    0
    I need help, but it is not homework for school. It is homework for myself.

    What I am doing now is making a hardwired ROM out of discrete logic gates, and part of my problem is that I am lacking enough knowledge of demorgans theorem.

    I know the basics, but when I came to this site:
    http://www.generalnumbers.com/demorgans_application1.html

    It confused me a bit. It states that

    AB + CD = ((AB)' (CD)')'

    Now It is easy to prove it by drawing the logic gates out on paper.
    But there has to be a way to prove it equation wise.

    If so, can someone explain it to me without skipping a step?
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,136
    1,786
  3. Tails

    New Member

    Feb 13, 2007
    5
    0
    De Morgan's law states that: (X+Y+Z)'= X'Y'Z', (X'+Y'+Z')= XYZ,







    heres how to solve this:


    AB+CD=((AB)'+(CD)')'

    we use De Morgans Law so that we can distribute the outside prime


    AB+CD=(A'+B')'+(C'+D')'

    then we apply De Morgans law again so that

    AB+CD= (AB)+(CD)


    hope that helps!!!
     
  4. mogadeet

    Member

    May 1, 2007
    19
    0
    hey mik

    here is a proof of your equation:

    ((AB)' (CD)')'

    = ((AB)'' + (CD)'') [by demorgan's theorem]

    = ((AB) + (CD)) [by cancellation of double negatives]


    for demorgan's theorem itself note that the following are all equivalent:

    (XY)' is true
    XY is false
    one of X,Y is false
    one of X',Y' is true
    X'+Y' is true

    peace
    stm
     
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