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The classical Laver table $A_{n}$ is the unique algebraic structure $$(\{1,\dots,2^{n}\},*_{n})$$ such that $x*_{n}1=x+1\mod 2^{n}$ and $$x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$$ for all $x,y,z\in\{1,\dots,2^{n}\}$. Then there is a unique operation $\circ_{n}$ on $A_{n}$ such that $(A_{n},\circ_{n},2^{n})$ is a monoid and where $x\circ_{n}y=(x*_{n}y)\circ_{n}x$ for all $x,y\in A_{n}$.

Suppose now that $r,s,u$ are terms in the language $*,\circ$ with $r,s$ unary and with $u$ binary and where $(A_{1},*_{n},\circ_{n})\models r(1)=1,s(1)=1$. Then does there exist an $n$ along with $x,y\in A_{n}$ where

  1. $r(x)*_{n}x=s(y)*_{n}y$

  2. $u(x,y)\neq 2^{n}$, and

  3. $\gcd(x,2^{n})=\gcd(y,2^{n})$?

The above result does hold in the algebras of rank-into-rank embeddings, but I do not know if it holds for the classical Laver tables.

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  • $\begingroup$ Under large cardinal hypotheses, the answer is affirmative if one removes condition 3. If $r$ is a term, then define a term $r^{\sharp}(x)=r(x)*x$. Then $$s^{\sharp}\circ t^{\sharp}=(t\circ s)^{\sharp}$$, and therefore $$s^{\sharp}\circ t^{\sharp}=(t\circ s)^{\sharp}=((t*s)\circ t)^{\sharp}=t^{\sharp}\circ(t*s)^{\sharp}.$$ So, if $x=s^{\sharp}(a),y=(s*r)^{\sharp}(a)$, then $r^{\sharp}(x)=s^{\sharp}(y)$ and with large cardinals, one can make $u(x,y)\neq 2^{n}$. The problem with this example is that $$\gcd(2^{n},y)=\gcd(2^{n},a)<\gcd(2^{n},x).$$ $\endgroup$
    – Joseph Van Name
    Commented Mar 8, 2019 at 17:52

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