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Suppose we have two binary operations $+$ and $*$. Suppose they are free objects in the variety of magmas. That is, the only equations they satisfy are when both terms on each side are identical terms. However, suppose that $*$ but not $+$ satisfy an additional constraint, which is that if $x*y$=$z*w$, then $x=z$ and $y=w$. What kind of name is used for this kind of operation? I thought it was "free object in the quasivariety of magmas", but I am not sure.

user107952
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    It’s no longer free when you add that constraint, right? And your purpose is not very clear to me: what does the addition have to do with your question? It seems like you can ignore it since you ask only about $*$ - they don’t even seem to be operations on the same set (although I am sort of guessing you have that in mind) –  Jul 28 '20 at 20:06
  • Yes, you can just ignore it. – user107952 Jul 28 '20 at 21:20
  • You still haven’t answered the main question: is $*$ actually free or is it a free object modulo a relation? –  Jul 28 '20 at 21:32
  • @TokenToucan I mean this. The only equations that $$ satisfy are $t=t$, for a term $t$. But also, it satisfies the condition that if $xy=z*w$, then $x=y$ and also $z=w$. To give a concrete example, natural numbers with tetration satisfy the first condition, but not the second. – user107952 Jul 28 '20 at 22:09
  • First, there is no magma at all that work with your first condition: the terms $"(a)"$ and $"a"$ are distinct but satisfy $(a) = a$. Second, you still haven't written things out carefully: you cannot insist that something only satisfies X (this "free" business) and then also ask that it satisfies Y (the extra relation) unless, for instance, you know Y follows from X. Lastly, I am pretty sure that a correct version of your freeness statement (e.g. actual freeness) implies the second condition and so your example with tetrations can't be sound. –  Jul 29 '20 at 00:21
  • I don't mean to be abrasive but I can't tell if you actual mean free when you say free, or you're asking whether free (whichever definition) implies the extra relation you want, or if you're asking for a name for the second property. In any case, the probable answer is that there is no name. –  Jul 29 '20 at 00:24
  • @TokenToucan I apologize. I don't really know what free means. But what I mean by it, in the context of magmas, is that the operation does not satisfy any non-trivial equations. A term in that language, is something like $(xx)(y*z)$. Basically, we have a countably infinite set of variables, and we build terms in that language using that operation. So, I am saying the only equations that it satisfies are of the form $t=t$. I don't know if that is the definition of free, but regardless, that is what I am talking about. – user107952 Jul 29 '20 at 00:35

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