# Left Distributive and Commutative implies Distributive

## Theorem

Let $\left({S, \circ, *}\right)$ 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

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

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

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

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

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

$\blacksquare$