Prove that $x\cdot y=0 \iff x=0$ or $y=0$ without cases I have proved that given $x,y \in F$, $F$ a field, $x\cdot y=0 \iff x=0$ or $y=0$ by making cases for neither $x$ nor $y$ equals $0$ (and did a proof by contradiction) and then two cases for $x=0$ and $y=0$ (direct proofs).
As a matter of style, I'm not a big fan of either proofs by cases or proofs by contradiction.  Is there some slick way to avoid both and do a direct proof in 1 case that establishes this result?

Well, the statement $A\implies (B\lor C)$ is actually equivalent to the statement $$A\land \neg B\implies C$$
So you don't have to separate cases. You can prove that if $x\cdot y=0$ and $x\neq 0$, then $y=0$. By doing so, you prove the statement $$x\cdot y=0\implies x=0\lor y=0$$
The other direction is obvious.
