Suppose that $f,g:X^{2}\rightarrow X$, and $T:X^{2}\rightarrow X^{2}$ is the function where $T(x,y)=(f(x,y),g(x,y))$. Then $(X,f,g)$ is said to satisfy the Yang-Baxter equation if $(T\times 1_{X})\circ(1_{X}\times T)\circ(T\times 1_{X})=(1_{X}\times T)\circ(T\times 1_{X})\circ(1_{X}\times T)$. In other words, $(X,f,g)$ satisfies the Yang-Baxter equation precisely when it satisfies the following identities:

- $f(f(x,y),f(g(x,y),z))=f(x,f(y,z))$
- $g(f(x,y),f(g(x,y),z))=f(g(x,f(y,z)),g(y,z))$
- $g(g(x,y),z)=g(g(x,f(y,z)),g(y,z))$

The collection of all algebras that satisfy the Yang-Baxter equations is a variety. Is this variety generated by its finite members?

Does there exist an inverse system $((X_{n})_{n},(\phi_{m,n})_{m,n})$ of finite algebras that satisfy the Yang-Baxter identity along with a sequence $(e_{n})_{n}\in\varprojlim X_{n}$ where each $X_{n}$ is generated by $e_{n}$ and where $(e_{n})_{n}$ generates a free subalgebra of $\varprojlim X_{n}$?

The motivation of this question comes from the fact that the Yang-Baxter equation is a generalization of self-distributivity, and if $T(x,y)=(x*y,x)$, then $T$ satisfies the Yang-Baxter equation precisely when $T$ is self-distributive. The variety of self-distributive algebras is generated by its finite members (in particular, the multigenic Laver tables) and the free self-distributive algebra on one generator embeds into an inverse limit of finite self-distributive algebras known as the classical Laver tables.