Let $X$ be the class of all classical Laver tables. Is $HS(X)=S(X)$?

Question

Let $A_{n}=(\{1,\ldots,2^{n}\},*_{n})$ be the algebra defined by $x*_{n}1=x+1\mod 2^{n}$ and $x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$ for all $x,y,z\in\{1,\ldots,2^{n}\}$. Suppose that $X$ is a subalgebra of some $A_{n}$ and $\simeq$ is a congruence on $X$. Then does there necessarily exist some $m$ along with some embedding $i:(X/\simeq)\rightarrow A_{m}$?

If so, then for all $n$, let $t_{n}$ be the least natural number $m$ such that whenever $X$ is a subalgebra of $A_{n}$ and $\simeq$ is a congruence on $X$, then there is some embedding $i:X/\simeq\rightarrow A_{m}$. Then what are some bounds on $t_{n}$? Or what is the exact value of $t_{n}$?

Answer 1

It is known that if $X$ is a cyclic subalgebra of $A_n$ and $\simeq$ is a congruence on $X$, then $X/\simeq$ is isomorphic to $A_m$ for some $m\leq n$. See Propositions 1.6 and 1.21 of Chapter X of Braids and Self-Distributivity by Dehornoy for the proof.

I don't think it is known whether the result is true for arbitrary subalgebras $X\leq A_n$. I experimented with $A_3$ and $A_4$ using UACALC, and could not find any counterexamples.