# Attraction in Laver tables

If $$X$$ is a self-distributive algebra, then define $$x^{[n]}$$ for all $$n\geq 1$$ by letting $$x^{[1]}=x$$ and $$x^{[n+1]}=x*x^{[n]}$$. The motivation for this question comes from the following fact about self-distributivity on one generator.

Theorem: Suppose that $$X$$ is a self-distributive algebra generated by one element. Then for all $$x,y\in X$$, there are $$m,n$$ with $$x^{[m]}=y^{[n]}$$.

Define the Fibonacci terms by letting $$t_{1}(x,y)=y,t_{2}(x,y)=x,t_{n+2}(x,y)=t_{n+1}(x,y)*t_{n}(x,y)$$.

An algebra $$(X,*,1)$$ is said to be a reduced permutative algebra if $$x*(y*z)=(x*y)*(x*z),x*1=1,1*x=x$$ for all $$x,y,z\in X$$ and where for all $$x,y\in X$$, there is some $$n$$ where $$t_{n}(x,y)=1$$. We say that a reduced permutative algebra $$(X,*,1)$$ is critically simple if whenever $$\simeq$$ is a congruence, either $$\simeq$$ is the trivial congruence or $$x\simeq 1$$ for some $$x\neq 1$$.

Suppose that $$(X,*,1)$$ is a finite reduced permutative algebra and $$x,y\in X$$ and suppose that $$p\geq 0$$. Then does there a reduced permutative algebra $$(Y,*,1)$$, natural numbers $$m,n$$ and a homomorphism $$\phi:Y\rightarrow X$$ and $$x',y'\in Y$$ where $$\phi(y')=y,\phi(x')=x$$ and $$x'^{[m]}=y'^{[n]}$$ and where $$x'^{[m+p]}\neq 1$$? What if $$(X,*,1)$$ is critically simple?