Suppose that $(X,*)$ is an algebra that satisfies the self-distributivity identity $x*(y*z)=(x*y)*(x*z)$. We say that an element $x\in X$ is a left-identity if $x*y=y$ for all $x\in X$. Let $\mathrm{Li}(X)$ denote the collection of left identities in $(X,*)$. We say $L\subseteq X$ is a left-ideal if $x*y\in L$ whenever $y\in L$. Define the Fibonacci terms $t_{n}(x,y)$ for all $n\geq 1$ by letting $t_{1}(x,y)=y,t_{2}(x,y)=x$ and $$t_{n+2}(x,y)=t_{n+1}(x,y)*t_{n}(x,y)$$ for $n\geq 1$.
We say that an algebra $(X,*)$ that satisfies the self-distributivity is permutative if $\mathrm{Li}(X)$ is a left-ideal and if for all $x,y\in X$, there is some $n$ where $t_{n}(x,y)\in\mathrm{Li}(X)$. We say that an algebra $(X,*)$ is Laver-like if $\mathrm{Li}(X)$ is a left-ideal and whenever $x_{n}\in X$ for $n\in\omega$, there is some $N$ where $x_{0}*\dots*x_{N}\in\mathrm{Li}(X)$ (Implied parentheses are on the left $(a*b)*c=a*b*c$). Every Laver-like self-distributive algebra is permutative.
Define $x^{n}*y$ for all $n\geq 0$ inductively by letting $x^{0}*y=y,x^{n+1}*y=x*(x^{n}*y)$.
If $(X,*)$ is permutative, then define $\mathrm{crit}(x)\leq\mathrm{crit}(y)$ if there is some $n$ where $x^{n}*y\in\mathrm{Li}(X)$. If $(X,*)$ is permutative, then $\{\mathrm{crit}(x)\mid x\in X\}$ is a linearly ordered set where $\mathrm{crit}(z)=\max\{\mathrm{crit}(x)\mid x\in X\}$ if and only if $z\in\mathrm{Li}(X)$, and if $(X,*)$ is Laver-like, then $\{\mathrm{crit}(x)\mid x\in X\}$ is well-ordered.
Suppose that $\alpha$ is a countable ordinal. Then does there exist an algebra $(X,*,t)$ generated by a single element that satisfies the distributivity identity $x*t(y,z)=t(x*y,x*z)$ and where $(X,*)$ is Laver-like with $\{\mathrm{crit}(x)\mid x\in X\}$ order isomorphic to $\alpha+1$?
Suppose that $Q$ is a countable totally ordered set with a least and greatest element. Then does there exist an algebra $(X,*,t)$ generated by a single element that satisfies the identity $x*t(y,z)=t(x*y,x*z)$ and where $(X,*)$ is permutative with $\{\mathrm{crit}(x)\mid x\in X\}$ order isomorphic to $Q$?
