# 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

\(\displaystyle \paren {a * b} \circ c\) | \(=\) | \(\displaystyle c \circ \paren {a * b}\) | $\circ$ is commutative | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {c \circ a} * \paren {c \circ b}\) | $\circ$ is left distributive over $*$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a \circ c} * \paren {b \circ c}\) | $\circ$ is commutative |

So $\circ$ is right distributive over $*$.

Since $\circ$ is both left distributive and right distributive over $*$, it is distributive over $*$.

$\blacksquare$