Suppose that $(X,*,1)$ satisfies the identities
$x*(y*z)=(x*y)*(x*z),x*1=1,1*x=x$. Then we say that $(X,*,1)$ is a reduced Laver-like algebra if whenever $x_{n}\in X$ for all $n\in\omega$, there is some $N\in\omega$ where $x_{0}*\dots*x_{N}=1$ (we group the invisible parentheses on the left: $a*b*c=(a*b)*c$).

Suppose that $((X_{n})_{n\in\omega},(\phi_{n,m})_{n\geq m})$ is an inverse system of Laver-like algebras where each transitional mapping $\phi_{n,m}$ is surjective. Let $A$ be a finite set. Suppose that for each $n$, the algebra $X_{n}$ is generated by $(x_{n,a})_{a\in A}$ and $\phi_{n,m}(x_{n,a})=x_{m,a}$ whenever $a\in A,n\geq m$. Suppose furthermore that whenever $a_{1},\dots,a_{r}\in A$ there is some $n$ where
$x_{n,a_{1}}*\dots*x_{n,a_{r}}\neq 1$. 

If $m\in\omega,x,y\in X_{m}$, then do there exist $n\geq m$ along with $x',y'\in X_{n}$ where $\phi_{n,m}(x')=x,\phi_{n,m}(y')=y$ but where
$x'*y'\neq 1$?

If $|A|=1$, then the answer to the above question is yes under the hypotheses that for all $n$, there exists some $n$-huge cardinal. I strongly suspect that the answer is negative for $|A|>1$.