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Without De Morgan's

  1. Sep 1, 2011 #1
    Can we prove :([itex]A\cap B[/itex])' = [itex]A^'\cup B^'[/itex] ,without using De Morgan's??
     
  2. jcsd
  3. Sep 1, 2011 #2

    phinds

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    you can if you accept a Karnaugh map or a Venn diagram as proof
     
  4. Sep 1, 2011 #3
    no ,i mean an ordinary proof.
     
  5. Sep 1, 2011 #4

    phinds

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    well, try googling "proof of DeMorgan's Theorem" (or Law as it seems to be called these days)
     
  6. Sep 1, 2011 #5

    HallsofIvy

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    Just use the basic definitions:

    If [itex]x\in \left(A\cap B\right)'[/itex], then x is NOT in[itex]A\cap B[/itex] when means it is either not in A or not in B. If it is not in A then it is in A' and therefore in [itex]A'\cup B'[/itex]. If it is not in B then it is in B' and therefore in [itex]A'\cup B'[/itex]. Thus, [itex]\left(A\cap B\right)'\subset A'\cup B'[/itex].

    The other way: if [itex]x \in A' \cup B'[/itex] it is in either A' or in B'. If it is in A', ...
     
  7. Sep 1, 2011 #6

    That is using De Morgan's. But i asked for a proof without using De Morgans
     
  8. Sep 1, 2011 #7
    I am afraid i could not find a proof ,apart from one using truth tables
     
  9. Sep 1, 2011 #8

    phinds

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    And what does that tell you?
     
  10. Sep 1, 2011 #9

    rbj

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    how is De Morgan's theorem proven? you can restate the proof that uses De Morgan and where you get to the spot where De Morgan is invoked, then proceed with the steps that prove De Morgan but with your specific conditions (or "input") instead of the general A and B case.
     
  11. Sep 2, 2011 #10

    HallsofIvy

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    Then I am confused as to what you mean by "using DeMorgan's". What I gave is how one would prove DeMorgan's laws but did not use DeMorgan's laws.
     
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