Critique my proof of: P → (Q → R). Then ¬R → (P → ¬Q) I'm a beginner at mathematical proof and I want to improve my proof writing skills. Any critique on correctness, structure, etc. is much appreciated. Below is a theorem and my proof of it.
Theorem. Suppose $P → (Q → R).$ Then $¬R → (P → ¬Q).$
Proof. Suppose $¬R,$ $P,$ and $Q → R.$ Because $Q → R$ and $¬R,$ it must be the case that $¬Q$ because $P$ and $P → (Q → R).$ Therefore, because when $¬R,$ $P$ and $¬Q,$ it must be the case that $¬R→(P→ ¬Q).$
 A: 
Proof. Suppose $¬R,$ $P,$ and $Q → R.$ Because $Q → R$ and $¬R,$ it must be the case that $¬Q$ because $P$ and $P → (Q → R).$ Therefore, because when $¬R,$ $P$ and $¬Q,$ it must be the case that $¬R→(P→ ¬Q).$


Any critique on correctness, structure, etc. is much appreciated.

Yes! I am glad you are asking for feedback on your proof, rather than us giving you some kind of alternative proof. As you probably are well aware, you learn much more from trying to create your own proof and getting some feedback on it as opposed to receiving someone else's proof.
Accordingly, here is some feedback on your proof. First, a line by line analysis:

Suppose $¬R,$ $P,$ and $Q → R.$

OK, that's weird. Given that you need to show $¬R→(P→ ¬Q)$, it makes sense to suppose both $\neg R$ and $P$, but why would you suppose $Q → R$?
Maybe you saw that you can get $Q → R$ once you assume $P$ and given that you have the premise $P → (Q → R)$. But that means that you infer $Q → R$, rather than suppose it.
So, you should really stick to supposing just $\neg R$ and $P$. And, it might be good to also spell out what it is that you are now looking for (which is $\neg Q$), before moving on.

Because $Q → R$ and $¬R,$ it must be the case that $¬Q$ because $P$ and $P → (Q → R).$

This is confusing. It is, all by itself, true that "Because $Q → R$ and $¬R,$ it must be the case that $¬Q$". So, why would you follow this up by "because $P$ and $P → (Q → R).$"? The first inference does not depend on $P$ and $P → (Q → R)$, so don't use that second because.
Indeed, it seems that once again you are realizing that you can get $Q → R$ once you have assumed $P$ and given that you have the premise $P → (Q → R)$. So you should make that a stand-alone observation in your proof, rather than embedding that into some further inference.

Therefore, because when $¬R,$ $P$ and $¬Q,$ it must be the case that $¬R→(P→ ¬Q).$

This is very confusing too. You make it sound like you assume $¬R,$ $P$ and $¬Q$. But that is not the case. All you assumed were $\neg R$ and $P$. The $\neg Q$ you obtained as a result of that. And yes, you therefore get $¬R→(P→ ¬Q)$, but once again you should make that a separate sentence.
Here is what I would suggest is a corrected version of your proof:
Suppose $¬R$ and $P$. Because $P$ and premise $P → (Q → R)$, you get $Q → R$. Because $Q → R$ and $¬R,$ it must be the case that $¬Q$. Therefore, when you have $¬R$ and $P$ you get $¬Q$. So, it must be the case that $¬R→(P→ ¬Q).$
Note that this proof really expressed the same idea as your proof ... but expresses it much better. It more clearly expresses exactly what follows from what, and it more clearly breaks it down in separate steps. So this is something you will need to work on for your proofs.  Good luck!!
A: You can try a proof by contradiction.
Suppose only $P \to (Q \to R)$!
Now assume $\neg R$ and the negation of the consequent $P \to \neg Q$ of the conditional, that is, $\neg (P \to \neg Q)$ or $P \land Q$. Now we have premises $\neg R$, $P$, and $Q$. Since $P$ holds, we have $Q \to R$. Now from $Q$, we have $R$. From $\neg R \land R$, we get a contradiction by the law of excluded middle. Therefore, $\neg R \to (P \to \neg Q)$.
A: Another approach is to write both propositions in conjunctive normal form.  The first one is
$$
P \implies (\lnot Q \implies R) \\
\lnot P \lor (Q \lor R) \\
\lnot P \lor Q \lor R
$$
The second one is
$$
\lnot R \implies (P \implies \lnot Q) \\
R \lor (\lnot P \lor Q) \\
R \lor \lnot P \lor Q \\
\lnot P \lor Q \lor  R
$$
In fact, we have shown that the two propositions are equivalent, which is more than what was required.
