B-Algebra is Quasigroup

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Theorem

Let $\left({X, \circ}\right)$ be a $B$-algebra.


Then $\left({X, \circ}\right)$ is a quasigroup.


Proof

By the definition of a quasigroup it must be shown that $\forall x \in X$ the left and right regular representations $\lambda_x$ and $\rho_x$ are permutations on $X$.

As $\left({X, \circ}\right)$ is a magma:

$\forall x \in X$ the codomain of $\lambda_x$ and $\rho_x$ is $X$.

Hence it is sufficient to prove that $\lambda_x$ and $\rho_x$ are bijections.


We have that:

Therefore:

$\forall x \in X$: $\lambda_x$ and $\rho_x$ are injective mappings.


We also have that regular representations of $B$-algebras are surjective.

Therefore:

$\forall x \in X$: $\lambda_x$ and $\rho_x$ are both injective and surjective mappings.

Hence by definition are $\lambda_x$ and $\rho_x$ bijections for all $x \in X$.

$\blacksquare$


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