If $P \iff Q$ is true, why does any proof of $Q$ have to use the fact that $P$ is true?

Question

If $P \iff Q$ is true, does any proof of $Q$ need to use the fact that $P$ is true?

I am asking this because my course notes say that because a stable matching does not always exist with only one group, and any proof of a stable matching always existing must use the fact that there are two different groups of people where people of the the same group cannot match with each other. My understanding is that here, (there being two distinct groups of people to match) $\iff$ (there will always be a stable matching).

When trying to prove that a stable matching always exists, I am trying to see why we must use the condition that is sufficient and necessary for it. More generally, when $P \iff Q$ is true and trying to prove that $Q$ is true, why we must use the fact that $P$ is true? It seems that when $P \iff Q$ is true, we can prove $Q$ using the fact that $P$ is either true or false, and if $P$ is false then the principle of explosion can help us prove $Q,$ so we don't need to use the fact that it's true.

Answer 1

In the stable marriage problem, there is always a stable matching.

In the stable roommate problem, there isn't always a stable matching.

So any proof of the existence of a stable matching in the marriage situation has to assume something that is true of marriage, but false for roommates. Otherwise the marriage proof would also apply to the roommate proof. And so one has to use the assumptions that in marriage, the preferences form a bipartite graph. Or something weaker than that but that still distinguishes the two scenarios.

Something similar used to be common to quickly rule out bad proofs of Fermat's Last Theorem, solutions to $x^n + y^n = z^n$ do not exist iff $n > 2$. Those who would submit annoying amateur proofs would be asked "where do you use the assumption that $n > 2$?", and if they couldn't say, then the proof was rejected because it also would have been a proof of the nonexistence of pythagorean triples.

Answer 2

The statement
"$P$ is true only if $Q$ is true"
means that "$P$ implies $Q$".

To prove proposition $P$, it is not necessary to use the proposition $Q$.

Example:
$P:$ a sequence $s$ is convergent.
$Q:$ a sequence $s$ is bounded.
Now, in order to prove sequence $s$ is convergent, we do not need to use that it $s$ bounded.

On the other hand though, in order to prove $P$ is false, it is indeed sufficient to prove that $Q$ is false. If $s$ is indeed an unbounded sequence, then it is necessarily divergent.

So, the verification of proposition $P$ may or may not include the use of the implied proposition of $Q$, but the falsification of $Q$ falsifies $P$.

Answer 3

when $P \Leftrightarrow Q$ is true and trying to prove that $Q$ is true, why we must use the fact that $P$ is true?

Based on your sequence of edits in response to feedback, it appears that you misread the sentence Q⟺P as “$Q$ only if $P$”, which you then misread as meaning that proving $Q$'s truth requires assuming $P$'s truth. If indeed so, first check out this answer: What is the exact meaning of “only” in “only if”?.

(Fact: when given P⇔Q, deducing $Q$ by assuming $P$ is essentially and fallaciously begging the question.)

It seems that when $P \Leftrightarrow Q$ is true, we can prove $Q$ using the fact that $P$ is either true or false, and if $P$ is false then the principle of explosion can help us prove $Q,$ so we don't need to use the fact that it's true.

No. If $P$ is false, the principle of explosion merely says that $$P\Rightarrow Q$$ is true, not that $$Q$$ is true. (The word “implication” informally sometimes refers to its consequent, so a common mistake is to conflate an implication's truth and its consequent's truth.)

(Fact: when given ¬P and P⇔Q, it is impossible to prove $Q.$)