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Let $ A_{n}=(\{1,\ldots,2^{n}\},*_{n})$ be the algebra defined by $ 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,\ldots,2^{n}\}$ . Suppose that $ X$ is a subalgebra of some $ A_{n}$ and $ \simeq$ is a congruence on $ X$ . Then does there necessarily exist some $ m$ along with some embedding $ i:(X/\simeq)\rightarrow A_{m}$Read more