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Thread: De Morgan's laws

  1. #1
    Super Member Showcase_22's Avatar
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    De Morgan's laws

    The set (A\B)\C is equal for all sets A,B and C to one of the following sets:

    i). A\(B $\displaystyle \cap$ C)
    ii). A\(B\C)
    iii). A\(B $\displaystyle \cup$ C)

    Which one?
    This question can be done using truth tables, but it would take a really long time!

    I was wondering if there was a way to do it using De Morgan's laws. This is what i've done so far:

    (A\B)\C=(A\B) $\displaystyle \cap C^c$=$\displaystyle (A \cap B^c)\cap C^c$

    I'm not sure where to go from here. I tried expanding the other expressions to see if I could get to this, but it hasn't worked.
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  2. #2
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    $\displaystyle (A \cap B^c)\cap C^c = A \cap (B^c \cap C^c) = A \cap (B \cup C)^c$

    Looks familiar, doesn't it?
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  3. #3
    Super Member Showcase_22's Avatar
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    $\displaystyle
    (A \cap B^c)\cap C^c = A \cap (B^c \cap C^c) = A \cap (B \cup C)^c
    $

    lol, it does (and is):

    $\displaystyle
    (A \cap B^c)\cap C^c = A \cap (B^c \cap C^c) = A \cap (B \cup C)^c
    $=A\(B $\displaystyle \cup$ C)

    So when you have an expression with all $\displaystyle \cup$'s or $\displaystyle \cap$'s you can change around the brackets? (ie. It's associative?)
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post
    So when you have an expression with all $\displaystyle \cup$'s or $\displaystyle \cap$'s you can change around the brackets? (ie. It's associative?)
    Yes, it's associative.
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