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This is related to a question I asked before. In that question, I asked if the meet of the commutative and associative properties can be axiomatized by the equation $(x * y) * x = x * (y * x)$. The answer was no. So, now I am asking, if we add also the equation, $((x*y)*z)*(x*(y*z))=(x*(y*z))*((x*y)*z)$ to the previous equation, would we get an axiomatization of the meet. It would be preferable if someone could give a finite axiomatization, or prove that there is none, but that is optional for this particular question.

user107952
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1 Answers1

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Consider the algebras with the following multiplication tables:

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The first is commutative, but not associative, since $$(0 \cdot 0) \cdot 1 = 1 \cdot 1 = 0 \neq 1 = 0 \cdot 0 = 0 \cdot (0 \cdot 1).$$ On the other hand, being commutative, it trivially satisfies your equalities.

The second is the two-element left-zero semigroup, defined by $x\cdot y = x$.
So it trivially satisfies your equalities (since the left most variable is the same on both sides of bot equalities).
Clearly, it is not commutative.

Hence, those two equalities combined don't entail neither associativity nor commutativity.

amrsa
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  • I believe there is a misunderstanding. I am looking for an identity that follows from the commutative identity, and also follows from the associative identity, but does not follow from my two identities. – user107952 Sep 24 '20 at 18:41
  • @user107952 I think I see what you mean. We have different ideas about the order in the lattice of sub varieties of a variety. One is the order dual of the other, that is, I see it as your order, up side down. – amrsa Sep 24 '20 at 19:44
  • @amsra So, can you give me such an identity? – user107952 Sep 27 '20 at 19:53
  • @user107952 I don't think so. I don't even know that that variety is finitely based. I suggest you to ask it on mathoverflow, but to actually ask for the join (and not the meet) of the varieties of semigroups and commutative groupoids (or magmas) in the lattice of subvarieties of the variety of all algebras of type $2$. I suppose what you want is an equational base for that variety. – amrsa Sep 28 '20 at 08:52
  • @amrsa In fact, the OP did ask the question on Math Overflow back in July: https://mathoverflow.net/questions/365616/is-there-a-finite-equational-basis-for-the-join-of-the-commutative-and-associati. – Alex Kruckman Sep 28 '20 at 11:49
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    @AlexKruckman OK, I see, and indeed, before that here, following the link in the first line of the post you linked to. The OP seems to really want to know whether or not there is a finite equational basis for the variety. I don't know how to research that problem, at least with a reasonable effort to give a satisfying answer. But the question seems very reasonable to ask, and certainly someone though about it before... – amrsa Sep 28 '20 at 13:18