Is there an identity between the associative identity and the constant identity?

Question

This is a follow-up to my previous question, here: Is there an identity between the commutative identity and the constant identity?. Let our signature be that of a single binary operation $+$. I define the constant identity to be $x+y=z+w$. The associative identity is $(x+y)+z=x+(y+z)$. Is there an identity strictly between the two? Meaning, is there an identity $E$ such that the constant identity implies $E$ but not conversely, and $E$ implies the associative identity but not conversely? Since the answer to my previous question was affirmative, I suspect the answer to this question will be affirmative also. But I can't find such an identity.

Answer 1

An identity $E$ that obeys all the claimed properties is $$ E: x+(y+z) = (x+y)+w \hbox{ for all } x,y,z,w.$$

• $E$ is implied by triple constancy (and hence by constancy): obvious since both sides are constant in this case
• $E$ does not imply triple constancy (and hence does not imply constancy either): follows from considering the left-zero semigroups $x+y=x$ mentioned by arsmath
• $E$ implies associativity: obvious by specializing to $w = z$
• $E$ is not implied by associativity: follows from considering (say) addition on the integers

This candidate $E$ was located by pursuing the analysis in Pace's answer to isolate the form that $E$ had to take as much as possible, as described in the comments to that answer. With a little more effort, it should be possible to entirely classify (up to relabelings and symmetries) the full set of identities $E$ that answer the question.

Here is the Hasse diagram of the various identities discussed on this page and on the related question linked by the OP, where the ordering is from stronger identities to weaker ones:

It might be a suitable undergraduate research project to extend this diagram to cover other short identities for magmas. EDIT: It might be a suitable graduate research project to find a way to do this automatically using proof assistants and possibly also machine learning/AI tools. (Some further discussion of this latter possibility can be found here, where I pose a concrete challenge of using such tools combined with human expert-hours to extend the above Hasse diagram to the five thousand or so other universal equational laws for magmas that involve at most four applications of the binary operation $+$.)

UPDATE: I am now launching a collaborative project to expand this graph to a much larger set of equational theories of Magmas. The github repository of this project is here, and a blog post describing the project can be found here.

SECOND UPDATE: Thanks to the project, we can now describe the full list of intermediate laws of order at most 4:

The associative law (4512) is in the red box on top; the constant law (46) is in the blue box on the bottom (together with several equivalent laws), and all other laws displayed are answers to the OP's question. (A more interactive version of this image can be found here.)

Answer 2

This is a partial answer to YCor's improved question about an identity between associativity and constancy of triples, reducing it to a finite check. (I'll use multiplicative notation below, for simplicity.)

Consider the multiplication on the set $\{a,b,c,d,e,0\}$ where $ab=d$, $dc=e$, and all other products are $0$. This algebra satisfies any identity where both sides involve four or more letters (equivalently, 3 or more products) since such products are $0$. Moreover, this algebra is not associative, since $a(bc)\neq (ab)c$. So the identity we are looking for must have at least one side that involves no more than 3 letters.

Now consider the multiplication on the set $\{a,b,0\}$ where $aa=b$, and all other products are $0$. This algebra satisfies constancy of triples. Any identity where one side has three or more letters and the other side has fewer than three letters is not satisfied in this algebra. So the identity we are looking for must have at least one side involving exactly 3 letters and the other side must involve at least 3 letters OR both sides involve two letters or less.

I believe the second option can be handled very easily, but in any case it is a finite check. So, consider the first option. Write such an identity as $t=t'$, where $t$ involves 3 letters exactly. Thus, up to symmetry and renaming, $t$ is of the form $x(yz)$, or $x(yy)$, or $x(yx)$, or $x(xy)$, or $x(xx)$.

It is easy to check that all five options are always $0$ in the first algebra we constructed above. So, if $t'$ involves four or more letters, then $t=t'$ is satisfied in the first algebra we constructed, but associativity doesn't follow.

Thus, $t'$ also involves exactly 3 letters. In order to avoid $t=t'$ being again satisfied in the first algebra, we must have $t'$ of the form $(pq)r$, where $p,q,r$ are distinct variables. These variables needn't be distinct from $x,y,z$.

We are now reduced to a finite check.