Zero Product with Proper Zero Divisor is with Zero Divisor

Theorem
Let $\left({R, +, \circ}\right)$ be a ring.

Let $x \in R$ be a proper zero divisor of $R$.

Then:
 * $\left({x \backslash 0_R}\right) \land \left({x \circ y = 0_R}\right) \land \left({y \ne 0_R}\right) \implies y \backslash 0_R$

That is, if $x$ is a proper zero divisor, then whatever non-zero element you form the product with it by to get zero must itself be a zero divisor.

Proof
Follows directly from the definition of proper zero divisor.

If $y \ne 0_R$ and $x \circ y = 0_R$ and $x \in R^*$ (which is has to be if it's a proper zero divisor), then all the criteria of being a zero divisor are fulfilled by $y$.