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Math Help - [SOLVED] simplifying using set identities

  1. #1
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    [SOLVED] simplifying using set identities

    Please help me I have no idea on this one, I just get stuck after one step. Show that (A -B) \cup (B-A) = (A \cup B) - (A \cap B) using (a) the element method and (b) set algebra.

    So far I expanded this to (A \cap B^c) \cup (B \cap A^c) = (A \cup B) \cap (A \cap B)^c by alternate representation of set difference, but then I'm stuck! Can anyone else solve it? Is it even possible?
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  2. #2
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    \begin{gathered}<br />
  \left( {A \cap B^c } \right) \cup \left( {B \cap A^c } \right) \hfill \\<br />
   = \left[ {\left( {A \cap B^c } \right) \cup B} \right] \cap \left[ {\left( {A \cap B^c } \right) \cup A^c } \right] \hfill \\<br />
   = \left[ {A \cup B} \right] \cap \left[ {A^c  \cup B^c } \right] \hfill \\<br />
   = \left( {A \cup B} \right) \cap \left( {A \cap B} \right)^c  \hfill \\ <br />
\end{gathered}
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  3. #3
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    Thanks - that's very helpful, but how did you go from the first step to the second step (i.e. why does (A \cap B^c ) \cup (B \cap A^c ) = [(A \cap B^c ) \cup B] \cap [(A \cap B^c ) \cup A^c ])?
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  4. #4
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    Set Theory Proof

    Hello horan
    Quote Originally Posted by horan View Post
    Thanks - that's very helpful, but how did you go from the first step to the second step (i.e. why does (A \cap B^c ) \cup (B \cap A^c ) = [(A \cap B^c ) \cup B] \cap [(A \cap B^c ) \cup A^c ])?
    The Distributive Law:

    C \cup (B \cap A^c ) = [C \cup B] \cap [C  \cup A^c ]

    with C replaced by (A\cap B^c) gives

    (A \cap B^c ) \cup (B \cap A^c ) = [(A \cap B^c ) \cup B] \cap [(A \cap B^c ) \cup A^c ]

    Grandad
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