THM: $\displaystyle (A-B) - C = (A-C) - (B -C)\ \ \forall $ sets $\displaystyle A,B,C$. Prove the above thm.
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$\displaystyle \begin{array}{l} \left( {A \cap C^c } \right) \cap \left( {B \cap C^c } \right)^c \\ \left( {A \cap C^c } \right) \cap \left( {B^c \cup C} \right) \\ \left( {A \cap C^c \cap B^c } \right) \\ \end{array} $
Originally Posted by Plato $\displaystyle \begin{array}{l} \left( {A \cap C^c } \right) \cap \left( {B \cap C^c } \right)^c \\ \left( {A \cap C^c } \right) \cap \left( {B^c \cup C} \right) \\ \left( {A \cap C^c \cap B^c } \right) \\ \end{array} $ I appreciate it. I believe the $\displaystyle C^c$ for instance means the compliment of C (that is everything that is not C)- is there another way to prove this? I don't quite see how it proves it. Thanks for the explanation.
$\displaystyle \left( {A \cap C^c \cap B^c } \right) = \left( {A \cap B^c } \right) \cap C^c = \left( {A\backslash B} \right)\backslash C $
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