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

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 $$y$$.