Epimorphism Preserves Distributivity

Theorem
Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be algebraic structures.

Let $\phi: R_1 \to R_2$ be an epimorphism.


 * If $\circ_1$ is left distributive over $+_1$, then $\circ_2$ is left distributive over $+_2$.


 * If $\circ_1$ is right distributive over $+_1$, then $\circ_2$ is right distributive over $+_2$.

Consequently, if $\circ_1$ is distributive over $+_1$, then $\circ_2$ is distributive over $+_2$.

That is, epimorphism preserves distributivity.

Proof
Throughout the following, we assume the morphism property holds for $\phi$ for both operations.

It remains to be shown that $\paren {x +_1 y}, \paren {y +_1 z}, \paren {x \circ_1 y }, \paren {y \circ_1 z}, \paren {x \circ_1 z}, \paren { x \circ_1 \paren {y +_1 z} }, \paren { \paren {x +_1 y} \circ_1 z } \in \Dom \phi$.

Left Distributivity
Suppose $\circ_1$ is left distributive over $+_1$.

Then:

So $\circ_2$ is left distributive over $+_2$.

Right Distributivity
Suppose $\circ_1$ is right distributive over $+_1$. Then:

So $\circ_2$ is right distributive over $+_2$.

Distributivity
If $\circ_1$ is distributive over $+_1$, then it is both right and left distributive over $+_1$.

Hence from the above, $\circ_2$ is both right and left distributive over $+_2$.

That is, $\circ_2$ is distributive over $+_2$.