[Math] Prime Ideals and multiplicative sets

Question

I am currently studying a course on commutative algebra and came across this statement:

An Ideal $I$ in a ring $R$ is prime if and only if $R\setminus I$ is a multiplicative set.

I have proved it the following way and was wondering if anyone would be kind enough to go through the proof to check for any errors.

So assume $R\setminus I$ is multiplicative, then $f\in R\setminus I$, $g \in R\setminus I\implies fg \in R\setminus I$. This is the same as saying $f,g$ not in $I$ implies $f.g$ not in $I$. So we just take the contrapositive of this to get our implication that $f.g$ $\in$ $I$ $\implies$ $f$ $\in$ $I$ or $g$ $\in$ $I$. Hence I is prime.

Now suppose $I$ is prime. Then,

$f.g$ $\in$ $I$ $\implies$ $f$ $\in$ $I$ or $g$ $\in$ $I$. Again taking the contrapositive of this, we can deduce the relation from above which would show that $R\setminus I$ is indeed multiplicative.

Is this a valid enough proof or would I need to be more "rigourous"?

Answer 1

I correct one direction. can you correct the other?

I use Transposition: $P \implies Q \equiv (\lnot Q\implies \lnot P)$. (and De Morgan's Theorem)

Assume $R\setminus I$ is multiplicative set. So:
$$(f \in R\setminus I) \land (g\in R\setminus I)\implies f g \in R\setminus I.$$ This is the same as saying
$$f g \notin R\setminus I\implies (f\notin R\setminus I) \lor (g\notin R\setminus I)$$ This is the same as saying $$fg\in I\implies (f\in I) \lor (g\in I).$$ So $I$ is is prime.