A finite reduced Laver-like algebra is a finite algebra $(X,\*,1)$ that satisfies the identities $1\*x=x,x\*1=1,x\*(y\*z)=(x\*y)\*(x\*z)$ and where there is a natural number $n$ and a function $\mathrm{crit}:X\rightarrow n+1$ where


1. $\mathrm{crit}(x)=n$ if and only if $x=1$,
2. $\mathrm{crit}(x\*y)=\mathrm{crit}(y)$ whenever $\mathrm{crit}(y)<\mathrm{crit}(x)$, and
3. $\mathrm{crit}(x\*y)>\mathrm{crit}(y)$ whenever
$\mathrm{crit}(x)\leq\mathrm{crit}(y)<n$.


We say that a finite reduced Laver-like algebra $(X,\*,1)$ is critically simple if whenever $\simeq$ is a congruence on $(X,\*,1)$ where there are $x,y\in X$ with $x\simeq y,x\neq y$, there is some $z\in X\setminus\{1\}$ with $z\simeq 1$.


We say that $(X,\*,1)$ is critically subsimple if $x,y\in X,x\*x=1,y\*y=1,\mathrm{crit}(x)=\mathrm{crit}(y)$ implies that $x=y$ for each $x,y\in X$. Every critically simple finite reduced Laver-like algebra is critically subsimple.


Suppose that $(X,\*,1)$ is a critically subsimple finite reduced Laver-like algebra generated by $(x\_{a})\_{a\in A}$ where $A$ is finite. Then does there exist a critically simple finite reduced Laver-like algebra $(Y,\*,1)$ generated by $(y\_{a})\_{a\in A}$ along with a homomorphism $\phi:Y\rightarrow X$ such that $\phi(y\_{a})=x\_{a}$ for each $a\in A$ and an element $c\in Y$ such that if $y\_{1},y\_{2}\in Y$, then $\phi(y\_{1})=\phi(y\_{2})$ if and only if $c\*y\_{1}=c\*y\_{2}$?



Is every critically subsimple Laver-like algebra a quotient of a critically simple Laver-like algebra on the same number of generators?
No. A critically subsimple Laver-like algebra is not necessarily a quotient of a critically simple Laver-like algebra with the same number of generators and more critical points. Our strategy for producing counterexamples is to exhibit finite reduced Laver-like algebras $X$ generated by $(x\_a)\_{a\in A}$ such that there does not exist a finite reduced Laver-like algebra $Y$ generated by $(y\_a)\_{a\in A}$ such that $Y$ has more critical points than $X$ and such that there is a homomorphism $\phi:Y\rightarrow X$ with $\phi(y\_a)=x\_a$ for $a\in A$.


Given a finite reduced Laver-like algebra $X$ and a generating set $(x\_a)\_{a\in A}$, I have obtained an algorithm for finding the class $\text{Cov}(X,(x\_a)\_{a\in A})$ of all finite reduced Laver-like algebras $Y$ together with a generating set $(y\_a)\_{a\in A}$ such that $Y$ has $1$ more critical point than $X$ and where there is a homomorphism $\phi:Y\rightarrow X$ with $\phi(y\_a)=x\_a$ for $a\in A$.


Producing finite reduced Laver-like algebras where $\text{Cov}(X,(x\_a)\_{a\in A})=\emptyset$ is not trivial since examples of algebras $(X,(x\_a)\_{a\in A})$ with $\text{Cov}(X,(x\_a)\_{a\in A})=\emptyset$ are difficult to obtain, and I have not been able to obtain any algebras $(X,(x\_a)\_{a\in A})$ with $\text{Cov}(X,(x\_a)\_{a\in A})=\emptyset$ without explicitly looking for such an algebra. I only know of one way of constructing such an algebra.


The way I constructed an algebra was to first perform a brute force search for a small reduced Laver-like algebra $(X,\*)$ generated by two elements $x,y$ with $y\*y=1$ and where if $\phi:Y\rightarrow X$ is a homomorphism between two Laver-like algebras and $Y$ is generated by $a,b$ and $\phi(a)=x,\phi(b)=y$, then $Y,X$ both have the same number of critical points. After, we have obtained the minimal Laver-like algebra $(X,\*)$, we produce a sequence $(X\_1,\dots,X\_n)$ where $X\_1=(X,(x,y))$, and $X\_{j+1}\in\text{Cov}(X\_j)$ for $1\leq j<n$, and where $\text{Cov}(X\_j)=\emptyset$. If say that two finite Laver-like algebras $X,Y$ are critically equivalent if there are congruences both denoted by $\simeq$ on $X,Y$ where $X/\simeq$ is isomorphic to $Y/\simeq$ and where $X,Y,X/\simeq,Y/\simeq$ all have the same number of critical points. We find some $Z$ that is critically equivalent to $X\_n$ but which is not critically subsimple and then $Z$ is our desired counterexample.


