DeMorgan's theorem problem

Discussion in 'Homework Help' started by mcc123pa, Sep 12, 2010.

  1. mcc123pa

    Thread Starter Member

    Sep 12, 2010
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    Hi-

    I was assigned this problem for homework:

    Using DeMorgan's theorem, express the function:
    F= AB'C+A'C'+AB
    a) with only OR and complement operations
    b) with only AND and complement operations

    I solved this problem and got the following answers:

    Part a:
    F=AB'C+A'C'+AB
    F=(AB'C)'+(A'C')'+(AB)'
    F=(A'+B+C')+(A+C)+(A'B')

    Part b:
    F=AB'C+A'C'+AB
    F=(AB'C)'+(A'C')'+(AB)'
    F=(A'.B.C').(A.C).(A'.B'.) (with the period being the AND function)

    Should I have double negated these at the first line?
    Are these answers correct or should I place a bar at the end of each parentheses thus negating everything inside of it once again?
     
  2. Georacer

    Moderator

    Nov 25, 2009
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    1,266
    Sadly they are wrong. Thing of it this way:
    If F=A+B+C, then certainly F isn't equivalent to F=A'+B'+C'! What you must do is this:

    F=(F')' %Double negation leaves the expression as is
    F=((A+B+C)')'
    F=(A'B'C')'

    That transforms the expression for use only with AND operators. For OR operators, just double negate the terms themselves. Am I clear?
     
  3. mcc123pa

    Thread Starter Member

    Sep 12, 2010
    40
    0
    Hi-

    Thanks for all of your help today so far!! I think I understand what you're saying, take a look at what I have here:

    F=AB'C+A'C'+AB

    part a)
    F=(F')'
    F=((AB'C+A'C'+AB)')
    F=(A'BC'.AC.A'B')' Is this how the final answer should look? Please let me know if it is worng.

    part b)
    F=(F')'
    F=A''+B''+A'''+C'''+A''+B''
    F=A+B+A'+C'+A+B Is this how the final answer should look for part B? Please let me know if it is wrong.
     
  4. Georacer

    Moderator

    Nov 25, 2009
    5,142
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    Well, not exactly...

    Here 's how it goes:

    part a):
    F=AB'C+A'C'+AB
    F=((AB'C)')'+((A'C')')'+((AB)')'
    ...

    part b):
    F=((AB'C+A'C'+AB)')'
    F=((AB'C)'(A'C')'(AB)')'
    ...
    Remember, (AB)'=A'+B' and (A+B)'=(A'B'). Can you continue from where I stopped?

    Also, what you wrote on part b) is totally wrong. Stick to your textbook rules and revise them frequently during your first steps of boolean algebra. It can get quite confusing.
     
  5. mcc123pa

    Thread Starter Member

    Sep 12, 2010
    40
    0
    Hi:

    Thanks for your response.

    I am still having difficulty with this problem.

    Part A:

    I took it from ((AB'C)')'+((A'C')')'+((AB)')'

    to (A'BC')'+(AC)'+(A'B')' and am stuck with how to procede.

    Part B:
    Unfortunately I am completely lost on this one.

    If you are able, would you be able to maybe give the answer and try to explain how you got there? I am lost unfortunately and would greatly appreciate it if you could do this. Thanks.
     
  6. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    OK, first of all let's clarify one thing:
    (AB'C)' equals A'+B+C'. Take a moment to understand why, taking on account my previous post (the "Remember" part).

    So we have:

    part a)
    F=AB'C+A'C'+AB
    F=((AB'C)')'+((A'C')')'+((AB)')'
    F=(A'+B+C')'+(A+C)'+(A'+B')'

    And there you have it, only OR and Complementations.

    part b)
    F=(F')'
    F=((AB'C+A'C'+AB)')'
    F=((AB'C)'(A'C')'(AB)')'

    That was ready since the last post, sorry for the misconception. As you can see, it contains only AND and Complementatry operators.

    Got it?
     
    Last edited: Sep 12, 2010
  7. mcc123pa

    Thread Starter Member

    Sep 12, 2010
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    0
    Yes, I understand now. Thanks so much for all of your help!
     
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