Can I assume a condition in the consequent? Im reading Axler's Linear Algebra Done Right. In an exercise, he ask to prove that $$a\in F,v\in V,av=0 \implies a=0 \lor v=0 $$ where $V$ is a vector space over the field $F$. I've proved it this way: First assume $av=0$. In order to prove that $a=0$, assume $v\neq 0$, then we have $(a+0)v=av+0v=0$. From here we conclude that $0v=-av$, and thus $a=0$. Now assume $a\neq 0$ to prove that $v=0$. We have $a(v+0)=av+a0=0$ and thus $v$ must be $0$ Is this proof correct?. I mean, can I assume $v\neq 0$ and $a\neq 0$ being $a$ and $v$ in the consequent?

In order to prove $a=0\vee q=0$, yes, it suffices to assume $q\
eq0$ and conclude $a=0$.

_Alternatively_ , it suffices to assume $a\
eq0$ and conclude $q=0$. You don't have to do both. (In fact, you shouldn't do both, because it's confusing.) Feel free to pick whichever tactic seems more convenient.

That's the good news! The bad news is that your proof is not correct. In the first proof, you proceeded directly from $0v=-av$ to $a=0$, but it isn't clear how. In the second proof, you proceeded directly from $av+a0=0$ to $v=0$, but again it isn't clear how. It seems as if you're assuming the statement you're trying to prove.

As a hint, think about $a^{-1}$.