More than One Right Zero then No Left Zero

Theorem
Let $\left({S, \circ}\right)$ be an algebraic structure.

If $\left({S, \circ}\right)$ has more than one left zero, then it has no right zero.

Likewise, if $\left({S, \circ}\right)$ has more than one right zero, then it has no left zero.

Proof

 * Let $\left({S, \circ}\right)$ be an algebraic structure with more than one left zero.

Take any two of these, and call them $z_{L_1}$ and $z_{L_2}$, where $z_{L_1} \ne z_{L_2}$.

Suppose $\left({S, \circ}\right)$ has a right zero. Call it $z_R$.

Then, by the behaviour of $z_R$, $z_{L_1}$ and $z_{L_2}$:


 * $z_{L_1} = z_{L_1} \circ z_R = z_R$
 * $z_{L_2} = z_{L_2} \circ z_R = z_R$

So $z_{L_1} = z_R = z_{L_2}$, which contradicts the supposition that $z_{L_1}$ and $z_{L_2}$ are different.

Therefore, in an algebraic structure with more than one left zero, there can be no right zero.

The same argument can be applied to an algebraic structure with more than one right zero.