Left Distributive and Commutative implies Distributive
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Theorem
Let $\struct {S, \circ, *}$ be an algebraic structure.
Let the operation $\circ$ be left distributive over the operation $*$.
Let $\circ$ be commutative.
Then $\circ$ is distributive over $*$.
Proof
Let $a, b, c \in S$.
Then
\(\ds \paren {a * b} \circ c\) | \(=\) | \(\ds c \circ \paren {a * b}\) | $\circ$ is commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {c \circ a} * \paren {c \circ b}\) | $\circ$ is left distributive over $*$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \circ c} * \paren {b \circ c}\) | $\circ$ is commutative |
So $\circ$ is right distributive over $*$.
Since $\circ$ is both left distributive and right distributive over $*$, it is distributive over $*$.
$\blacksquare$