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Is $A' \cap C'$ the same thing as $(A \cap C)'$ ? Here ' means the complementary set.

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    $\begingroup$ What do you mean by $A'$? $\endgroup$ Sep 11, 2015 at 15:35
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    $\begingroup$ DRAW A PICTURE, Draw a box with two overlapping circles in it, start there! $\endgroup$
    – Alec Teal
    Sep 11, 2015 at 15:35
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    $\begingroup$ Complement of $A$ @TimRaczkowski $\endgroup$
    – Alec Teal
    Sep 11, 2015 at 15:35
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    $\begingroup$ Can you come up with a counter-example? $\endgroup$
    – Raj
    Sep 11, 2015 at 15:38

5 Answers 5

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Let $\Omega = \{1, 2, 3\}, A = \{2, 3\}, C = \{3\}$. Now calculate $A' \cap C'$ and $(A \cap C)'$. What does that tell you?

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Almost, but you have to switch the union/intersection when you negate it. (I don't know how to make the symbols!)

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$A'\cap C'$ is the same as $(A\cup C)'$

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No, using De Morgan's laws you can show that (A ∩ C)' = (A' ∪ C')

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These two sets are not equal, an element from $A$ which doesn't belong to $B$ is in $(A\cap B)'$ but not in $A'\cap B'$.

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