Is it possible to use less than n^3 (where n is the cardinality of the set the operation works on)? That is, is there some way to choose triples to test for associativity so that the associativity of the triples that have not been chosen is implied by the ones that have been chosen?
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2If not, we could never prove $\Bbb R$ and $\Bbb Z$ are associative rings – pancini Feb 26 '22 at 05:59
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@pancini One might argue you're still checking every possibility, you're just checking a lot of them at once. Or, in the case of $\Bbb Z$ (or especially $\Bbb N$ with, say, Peano arithmetic) there isn't even much to argue if you're doing induction: then you are clearly checking every single triple one by one. – Arthur Feb 26 '22 at 06:56
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@Arthur true. It's kind of a philosophical debate though. Similar to the question of whether ZF has $9$ axioms or infinitely many. – pancini Feb 26 '22 at 07:00
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1@pancini That being said, I think this question is really asking "Is there an algorithm for determining whether a given (closed) binary operation on a set of $n$ elements is associative that's better (in some rigorous sense) than brute-force checking all $n^3$ possible associations?" or maybe even more specifically "Given a (closed) binary operation on a set of $n$ elements, is there a way to choose $n^3-1$ associations such that checking all of them implies the last association also associates?" – Arthur Feb 26 '22 at 07:03
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See the various answers (and their comments) to Associativity test for a Magma. – Dave L. Renfro Feb 26 '22 at 07:47