What historical event does the following rebus represent? $$ \therefore \forall A, B: (A \cap B = \emptyset) \implies (A \neq B) $$
Hint:
$\forall x \in S, \text{ where } |S| = 9.$ This isn't a rebus, but it describes one aspect of the event.
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1$\begingroup$ I believe it is $LaTeX$ rebus, not math haha. $\endgroup$– newzadMay 20, 2016 at 20:11
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2$\begingroup$ this is false, let $A=B=\emptyset$ $\endgroup$– JMPMay 20, 2016 at 20:53
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$\begingroup$ Your hint is not a fully formed mathematical statement. $\endgroup$– Deusovi ♦May 20, 2016 at 21:57
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$\begingroup$ @Deusovi I know, unfortunately. Couldn't think of a better way to state it. $\endgroup$– DylanSpMay 20, 2016 at 21:59
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1$\begingroup$ @Leppy: It's "an" if you don't pronounce the H. Some accents pronounce it, some don't. $\endgroup$– Deusovi ♦May 21, 2016 at 2:17
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2 Answers
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My guess is:
Brown v. Board of Education
Explanation:
Brown v. Board of Education refuted the idea of the "separate, but equal" doctrine. The main statement is that if sets A and B are disjoint, then it's implied A is not equal to B. Another way to describe disjoint sets is separate. So in other words, if A and B are separate, they are NOT equal.
As for the hint (credit goes to pacoverflow for figuring this out):
Brown v. Board of Education was a unanimous 9-0 decision and there are 9 elements in set S (the set of Supreme Court members with concurring opinions).
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$\begingroup$ Very close, but it's a more specific event. You don't have the right explanation for the hint. $\endgroup$– DylanSpMay 20, 2016 at 20:34
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Hm. Well, a literal translation yields:
Therefore, all sets that don't intersect are not the same.
I'm not sure how to relate this to a common phrase yet.
answered May 20, 2016 at 20:31
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$\begingroup$ I think it should be all sets that
do intersect$\endgroup$– stackErrMay 20, 2016 at 21:48 -
$\begingroup$ No, the intersection is the empty set. This means they don't intersect. $\endgroup$ May 22, 2016 at 2:36