What historical event does the following rebus represent? $$ \therefore \forall A, B: (A \cap B = \emptyset) \implies (A \neq B) $$


$\forall x \in S, \text{ where } |S| = 9.$ This isn't a rebus, but it describes one aspect of the event.

  • 1
    $\begingroup$ I believe it is $LaTeX$ rebus, not math haha. $\endgroup$
    – newzad
    May 20, 2016 at 20:11
  • 2
    $\begingroup$ this is false, let $A=B=\emptyset$ $\endgroup$
    – JMP
    May 20, 2016 at 20:53
  • $\begingroup$ Your hint is not a fully formed mathematical statement. $\endgroup$
    – Deusovi
    May 20, 2016 at 21:57
  • $\begingroup$ @Deusovi I know, unfortunately. Couldn't think of a better way to state it. $\endgroup$
    – DylanSp
    May 20, 2016 at 21:59
  • 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

2 Answers 2


My guess is:

Brown v. Board of Education


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).

  • $\begingroup$ Very close, but it's a more specific event. You don't have the right explanation for the hint. $\endgroup$
    – DylanSp
    May 20, 2016 at 20:34
  • $\begingroup$ Sweet, took another shot at it. :) $\endgroup$
    – Tonkleton
    May 20, 2016 at 20:43
  • $\begingroup$ Nope, still don't have it. $\endgroup$
    – DylanSp
    May 20, 2016 at 20:44
  • $\begingroup$ Third time's a charm? :) $\endgroup$
    – Tonkleton
    May 20, 2016 at 20:55
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    $\begingroup$ It was a unanimous 9-0 decision. $\endgroup$ May 20, 2016 at 21:48

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.

  • $\begingroup$ I think it should be all sets that do intersect $\endgroup$
    – stackErr
    May 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

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