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Using basic operators, derive the additional operator intersection.

$R \cap S = ?$

I thought it might be: $(R \cup S) - ((R -S) \cup (S -R))$ but I also think there might be an easier way of doing it?

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    Well, you are taking what is in at least one of the two sets, and throwing away everything that is in exactly one of the two sets. You could, instead, take everything that is in one of them, and then throw away everything that is in *just* that one, no?2011-01-18
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    Hmm, that is worded very confusingly. I'm not sure I quite get you.2011-01-18
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    You could write it as $R-(R-S)$ or $S-(S-R)$.2011-01-18
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    $R\cup S$ is everything that is in at least one of $R$ and $S$. $R-S$ is everything that is in $R$ but not in $S$; $S-R$ is everything that is in $S$ but not in $R$, so $(R-S)\cup(S-R)$ is everything that is in *exactly one* of $R$ and $S$. So $R\cup S- ((R-S)\cup(S-R))$ is everything which is in at least one of the two sets, minus everything which is in exactly one of them. Now, what do you get if instead of doing that, you take everything which is in $R$, and you throw away everything which is in *just* $R$?2011-01-18
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    OK, I get it now. Thanks!2011-01-18

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You could write $R \cap S = R-(R-S)$ or $R \cap S = S-(S-R)$. As Arturo mentioned, you are looking at one set and throwing away everything that is just in that set.