Every Operation is Distributive over Left Operation

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Theorem

Let $\struct {S, \circ}$ be an algebraic structure.

Let $\gets$ denote the left operation on $S$.


Then $\circ$ is distributive over $\gets$.


Proof

\(\ds \forall a, b, c \in S: \, \) \(\ds a \circ \paren {b \gets c}\) \(=\) \(\ds a \circ b\) Definition of Left Operation
\(\ds \forall a, b, c \in S: \, \) \(\ds \paren {a \circ b} \gets \paren {a \circ c}\) \(=\) \(\ds a \circ b\) Definition of Left Operation
\(\ds \leadsto \ \ \) \(\ds \forall a, b, c \in S: \, \) \(\ds a \circ \paren {b \gets c}\) \(=\) \(\ds \paren {a \circ b} \gets \paren {a \circ c}\) from above
\(\ds \forall a, b, c \in S: \, \) \(\ds \paren {a \gets b} \circ c\) \(=\) \(\ds a \circ c\) Definition of Left Operation
\(\ds \forall a, b, c \in S: \, \) \(\ds \paren {a \circ c} \gets \paren {b \circ c}\) \(=\) \(\ds a \circ c\) Definition of Left Operation
\(\ds \leadsto \ \ \) \(\ds \forall a, b, c \in S: \, \) \(\ds \paren {a \gets b} \circ c\) \(=\) \(\ds \paren {a \circ c} \gets \paren {b \circ c}\) from above

Hence the result by definition of distributive.

$\blacksquare$


Sources

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