# [SOLVED] simplifying using set identities

• Mar 4th 2009, 10:17 PM
horan
[SOLVED] simplifying using set identities
Please help me I have no idea on this one, I just get stuck after one step. Show that $\displaystyle (A -B) \cup (B-A) = (A \cup B) - (A \cap B)$ using (a) the element method and (b) set algebra.

So far I expanded this to $\displaystyle (A \cap B^c) \cup (B \cap A^c) = (A \cup B) \cap (A \cap B)^c$ by alternate representation of set difference, but then I'm stuck! Can anyone else solve it? Is it even possible?
• Mar 5th 2009, 03:04 AM
Plato
$\displaystyle \begin{gathered} \left( {A \cap B^c } \right) \cup \left( {B \cap A^c } \right) \hfill \\ = \left[ {\left( {A \cap B^c } \right) \cup B} \right] \cap \left[ {\left( {A \cap B^c } \right) \cup A^c } \right] \hfill \\ = \left[ {A \cup B} \right] \cap \left[ {A^c \cup B^c } \right] \hfill \\ = \left( {A \cup B} \right) \cap \left( {A \cap B} \right)^c \hfill \\ \end{gathered}$
• Mar 5th 2009, 08:26 AM
horan
Thanks - that's very helpful, but how did you go from the first step to the second step (i.e. why does $\displaystyle (A \cap B^c ) \cup (B \cap A^c ) = [(A \cap B^c ) \cup B] \cap [(A \cap B^c ) \cup A^c ]$)?
• Mar 6th 2009, 01:26 AM
Set Theory Proof
Hello horan
Quote:

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

The Distributive Law:

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

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

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