Is it always necessary to prove the ‘iff’ in both directions

Question

I have an exercise in my course, which asks to prove $A \cup B = B \iff A \subseteq B$.

My proof is: Let $A \nsubseteq B$, that is, $\exists a \in A : a \notin B$. Then from the definition follows $a \in A \cup B = B$, in contradiction to the initial assertion. $\square$

Usually I see that it's much more rigorous to prove $\implies$, then $\impliedby$, but I'm not sure, if that's only an option or a strict rule — and specifically if my proof does the job in both directions or there are some gaps that I don't recognize. My script suggests a really long 10+ lines proof using the 'both directions style', but I myself don't really see this necessity at least here.

This being said, is it always a must to prove the 'iff' in both directions?

Answer 1

It appears that you're trying (without making it completely clear) to prove $A\cup B=B \Leftrightarrow A\subseteq B$ by showing that $A\cup B=B$ together with $A\not\subseteq B$ leads to a contradiction.

If you think that is a complete proof, how about this one, by the same principle:

Claim: For any integer $n>2$, $$ n\text{ is prime} \iff n\text{ is odd} $$

Proof: Assume that $n$ is prime and that $n$ is not odd. Then $n$ is even, so $n=2k$ for some $k$. But then $2$ divides $n$, which is a contradiction with $n$ being prime, since $n>2$. $\Box$

This seems to follow exactly the same logic as your proof -- namely, considering $P\Leftrightarrow Q$ to be proved because I have shown that $\neg Q$ and $P$ together lead to a contradiction.

But there are odd numbers that are not prime -- such as $9$ -- so the claim is not actually true.