If $P$ is the only sufficient condition for $Q$ besides $Q$ itself, is $P$ necessary for $Q$?

Question

I am wondering whether if $P$ is the only sufficient condition for $Q$ besides $Q$ itself (by which I mean that $[P \implies Q] \wedge [(R \implies Q) \implies (R\equiv P \vee R \equiv Q)]$, then $P$ is necessary for $Q$. I am unsure because $Q$ is sufficient for $Q$, which makes it seem that $Q$ could be sufficient for itself and it could be true without $P$ being true, meaning that $P$ isn't necessary for $Q$.

Also, is is my reasoning correct for how I justify that for any proposition $Q$, $Q\implies Q$? I reason that because we assume that the same proposition has the same value in all contexts, the truth of $Q$ is sufficient to conclude the truth of $Q$ because $Q$ can't suddenly change into being false in the context in which we are considering its boolean value. Is this sound?

Answer 1

If $P$ is sufficient for $Q$, then trivially $P \land R$ is sufficient for $Q$ for each and every $R$.

So in exactly what good sense could $P$ be a sufficient condition for $Q$ and the only one (other than $Q$)?

Answer 2

I understand that by our hypothesis "$P$ is the only sufficient condition for $Q$ besides $Q$" it is meant that $$(P\rightarrow Q)\land \big((R\rightarrow Q)\rightarrow \neg(R\leftrightarrow Q)\rightarrow (P\leftrightarrow R)\big)$$

and that by our thesis "$P$ is necessary for $Q$" it is meant that $$\neg P\rightarrow\neg Q$$

The case that bothers us is when $P$ is false and $Q$ is true. Since it's the only one that makes the thesis false. We would like then in this case the hypothesis to be false as well.

Unfortunately even independent of $R$ the hypothesis becomes true. Which means your conjecture is not true.

Tell me if I misinterpreted. Hope this helps. Cheers!