(This is a follow-up question; [the original question](https://mathoverflow.net/questions/328269) was about *shelves*.)


A [rack](https://ncatlab.org/nlab/show/rack) $(R, \rhd, \lhd)$ is a set $R$ with two binary operations $\rhd$ and $\lhd$ such that for all $x, y, z \in R$:


1. $x \rhd (y \rhd z) = (x \rhd y) \rhd (x \rhd z)$;
2. $(x \lhd y) \lhd z = (x \lhd z) \lhd (y \lhd z)$;
3. $(x \rhd y) \lhd x = y = x \rhd (y \lhd x)$.


I am looking for a non-trivial rack $(R, \rhd, \lhd)$ such that for any two distinct $x, y \in R$:
$$
x \rhd y = y \iff y \rhd x \neq x.
$$
Can a rack with this property have more than one element?


(Or, in terms of [the previous question](https://mathoverflow.net/questions/328269): can the operation $\rhd$ of a non-trivial shelf with "trichotomy" have an inverse $\lhd$ in the sense of axioms (2) and (3) above?)



Racks with "trichotomy"
There are no such racks. We shall first show that quandles with multiple elements cannot satisfy the Trichotomy property. We shall then show that every rack that satisfy the Trichotomy property must be a quandle.


Suppose that $(Q,\*,\*^{-1})$ is a quandle ($\*$ is left-distributive i.e. $x\*(y\*z)=(x\*y)\*(x\*z)$). Then whenever $x\in G$, let $L\_{x}:Q\rightarrow Q$ be the inner automorphism defined by $L\_{x}(y)=x\*y$. Let $\mathrm{Inn}(Q)$ be the subgroup of $\mathrm{Aut}(Q)$ generated by $(L\_{x})\_{x\in X}$. Define a mapping $\phi:Q\rightarrow\mathrm{Inn}(Q)$ by letting $\phi(x)=L\_{x}$ for $x\in X$. Then $\phi$ is always a quandle homomorphism.


Suppose $Q$ satisfies the trichotomy property with $|Q|>1.$ Then the mapping $\phi$ is injective. If $x\neq y$ and $\phi(x)=\phi(y)$, then $y\*x=L\_{y}(x)=L\_{x}(x)=x\*x=x$ and $x\*y=y$ which contradicts the trichotomy property. Now, $x\*y=y$ if and only if $$\phi(y)=\phi(x\*y)=\phi(x)\*\phi(y)=\phi(x)\phi(y)\phi(x)^{-1}$$ if and only if $\phi(y)\phi(x)=\phi(x)\phi(y)$. Similarly, $y\*x=x$ if and only if $\phi(x)\phi(y)=\phi(y)\phi(x)$. Therefore, $x\*y=y$ if and only if $y\*x=x$ in contradiction of the trichotomy property.


We shall now show that if $X$ is a left shelf that satisfies the trichotomy property, then $t(a)=a\*a$ whenever $t$ is a unary term where $t(x)\neq x$ in the equational theory of shelves. As a corollary, if $(X,\*)$ is a rack ($X$ is left distributive) that satisfies the trichotomy property, then $x\*(x\*x)=x\*x$ which implies that $x\*x=x$ and therefore that $X$ is a quandle.


For left-shelves, define terms $x\_{[n]},x^{[n]}$ recursively by letting $x\_{[1]}=x=x^{[1]},x^{[n+1]}=x\*x^{[n]},x\_{[n+1]}=x\_{[n]}\*x$.


The classical Laver tables are the unique left shelves $A\_{n}=(\{1,\dots,2^{n}\},\*\_{n})$ where
$x\*\_{n}1=x+1\mod 2^{n}$ for $x\in A\_{n}$. If $n\geq 1,$ then the classical Laver table $A\_{n}$ does not satisfy the trichotomy property.


Proofs of the facts listed below can be found in Patrick Dehornoy's book Braids and Self-Distributivity (Chapters 5,10).


**Fact:** Suppose that $(X,\*)$ is a left-shelf generated by an element $e$ where $e\_{[2^{n}\cdot(2k+1)+1]}=e.$ Then there is some classical Laver table $A\_{m}$ for $m\leq n$ and an isomorphism $\phi:(X,\*)\rightarrow A\_{m}$ where $\phi(e)=1$.


**Fact:** If $(X,\*)$ is a left shelf generated by a single element $x$ and $a\in X.$ then there are $m,n,p>2$ where $a^{[m]}=x^{[n]}$ and where $a\*x^{[p]}=x^{[p+1]}$.


**Theorem:** Suppose that $(X,\*)$ is a left-shelf that satisfies the trichotomy property and $(X,\*)$ is generated by a single element $x$. Then $X=\{x,x\_{[2]}\}$ and $r\*s=x\_{[2]}$ whenever $r,s\in X$.


**Proof:** If $X\_{[2]}=x$, then $X=\{x\}$ and this proof is trivial. For this proof, let us assume that $x\_{[2]}\neq x$. We observe that no classical Laver table $A\_{n}$ for $n\geq 1$ satisfies the trichotomy property. Therefore, for all $n>1$, we have $x\_{[n]}\neq x$. Furthermore, since $x\neq x\_{[n+1]}=x\_{[n]}\*x$ and $x\_{[n]}\neq x$ for $n>1$, we conclude that $x\*x\_{[n]}=x\_{[n]}$ for $n>1$.


I now claim that $x^{[n]}=x^{[2]}$ for $n\geq 2$. Under the assumption that $x^{[n]}=x^{[2]}$, we have $$x^{[n+1]}=x\*x^{[n]}=x\*x^{[2]}=x^{[2]}$$ establishing our claim.


Suppose now that $a\in X$. Then there are $m,n,p>2$ where $$a^{[2]}=a^{[m]}=x^{[n]}=x^{[2]}$$ and where $$a\*x^{[2]}=a\*x^{[p]}=x^{[p+1]}=x^{[2]}.$$


We have, $$x\*x\_{[3]}=x\*(x\_{[2]}\*x)=(x\*x\_{[2]})\*(x\*x)=x\_{[2]}\*x\_{[2]}=x\_{[2]}.$$


Now, $x\_{[1]}\neq x\_{[3]},x\_{[3]}\*x\_{[1]}\neq x\_{[1]}$, so $x\_{[2]}=x\_{1}\*x\_{[3]}=x\_{[3]}$, so $x\_{[2]}=x\_{[3]}$.


Therefore, $$x\*x\_{[2]}=x\_{[2]},x\_{[2]}\*x\_{[2]}=x\_{[2]},x\_{[2]}\*x=x\_{[3]}=x\_{[2]},x\*x=x\_{[2]}.$$ Therefore, we conclude that $x,x\_{[2]}$ is a subalgebra of $X$, but since $X$ is generated by $x$, we conclude that $X=\{x,x\_{[2]}\}$ and $x\_{[2]}=r\*s$ whenever $r,s\in X$. **QED**


**Exercise:** I will therefore leave it as an exercise to rewrite this proof without referring to any facts about self-distributive algebras that I have referenced.


