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Math Help - sets prove

  1. #1
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    sets prove

    Show that for any sets A and B ,

    (AnB)'-(A'nB)=B'

    Thanks for helping ..
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by thereddevils View Post
    Show that for any sets A and B ,

    (AnB)'-(A'nB)=B'

    Thanks for helping ..
    For two sets P and Q, P-Q=P\cap Q'

    Thus (A\cap B)'-(A'\cap B)=(A\cap B)'\cap (A'\cap B)'

    By de Moivre's formula, this is :
    [(A\cap B)\cup (A'\cap B)]'

    And maybe it's clear for you that (A\cap B)\cup (A'\cap B)=B ?

    If you don't see it :
    - draw a diagram
    - or expand, and use the fact that A\cup A'=E, where E is the universe.
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  3. #3
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    Quote Originally Posted by thereddevils View Post
    Show that for any sets A and B ,
    (AnB)'-(A'nB)=B'
    \left( {A \cap B} \right)^c \backslash \left( {A^c  \cap B} \right) = \left( {A \cap B} \right)^c  \cap \left( {A^c  \cap B} \right)^c  = \left( {A^c  \cup B^c } \right) \cap \left( {A \cup B^c } \right)
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  4. #4
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    Hello, thereddevils!

    Show that for any sets A and B: . <br />
(A \cap B)'-(A' \cap B) \:=\:B'

    I'll signify a couple of axioms:

    . . (1)\;\;P \cap P' \:=\: \emptyset \qquad (2)\;\; P \cup \emptyset \:=\: P


    \begin{array}{ccccc}<br />
(A \cap B)' - (A' \cap B) & = & (A \cap B)' \cap (A' \cap B)' && \text{d{e}f. Subtraction} \\ \\<br /> <br />
& = & (A' \cup B') \cap (A \cup B') && \text{DeMorgan} \\ \\<br /> <br />
& = & (A' \cap A) \cup B' && \text{Distributive} \\ \\<br /> <br />
& = & \emptyset \cup B' && \text{axiom (1)} \\ \\<br /> <br />
& = & B' && \text{axiom (2)} \end{array}

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