Examples of nilpotent self-distributive algebras

Suppose that $ (X,*,1)$ is an algebra that satisfies the identities $ x*(y*z)=(x*y)*(x*z),x*1=1,1*x=x$ . Define the right powers by letting $ x^{[1]}=x$ and $ x^{[n+1]}=x*x^{[n]}$ . We say that $ (X,*,1)$ is a right nilpotent reduced self-distributive algebra if for each $ x\in X$ , there is an $ n>1$ where $ x^{[n]}=1$ . What are some examples of right nilpotent reduced self-distributive algebras $ (X,*,1)$ ?

Non-examples Racks, quandles, and spindles (spindles are self-distributive algebras $ (X,*)$ that satisfy the idempotence identity $ x*x=x$ ) with cardinality greater than $ 1$ are examples of self-distributive algebras that are never right nilpotent reduced self-distributive algebras. The right nilpotent reduced self-distributive algebras must therefore have a strong form of irreversibility.

Preliminary examples: Here are a few examples to get things started and to make this problem more fun.

  1. Suppose $ \lambda$ is a cardinal. If $ \mathcal{E}_{\lambda}$ is the set of all elementary embeddings $ j:V_{\lambda}\rightarrow V_{\lambda}$ , $ *$ is defined by $ j*k=\bigcup_{\alpha<\lambda}j(k|_{V_{\alpha}})$ $ \gamma$ is a limit ordinal with $ \gamma<\lambda$ and $ \equiv^{\gamma}$ is the congruence on $ \mathcal{E}_{\lambda}$ defined by $ j\equiv^{\gamma}k$ if and only if $ j(x)\cap V_{\gamma}=k(x)\cap V_{\gamma}$ for each $ x\in V_{\gamma}$ . Then, by the Kunen inconsistency, $ \mathcal{E}_{\lambda}/\equiv^{\gamma}$ is a right nilpotent reduced self-distributive algebra. This example can be generalized since there are finite algebras that resemble $ \mathcal{E}_{\lambda}/\equiv^{\gamma}$ but which cannot arise from the algebras of rank-into-rank embeddings.

  2. If $ \rightarrow$ is the Heyting operation in a Heyting algebra with maximum element $ 1$ , then $ (X,\rightarrow,1)$ is a right nilpotent reduced self-distributive algebra.

  3. Suppose that $ f:X\rightarrow X$ is a function such that $ f(1)=1$ and for all $ x\in X$ , there is an $ n$ where $ f^{n}(x)=1$ . Define $ *$ by letting $ x*y=f(y)$ . Then $ (X,*)$ is a right nilpotent reduced self-distributive algebra.