Prove that if P, then $Q → ¬(Q → ¬P)$ 
Suppose that $P$ is true. Prove that $Q → ¬(Q → ¬P)$ is true.

My attempt:
If $Q$, then statement $\lnot (Q → \lnot P)$ must be true. We know that $P$ is true. For $¬(Q → ¬P)$ to be true, $Q$ must be true. Therefore, given that $P$ is true, statement $Q → ¬(Q → ¬P)$ is true.
Is it correct?  Any suggestions to make it more concise/clear would be welcome.
 A: Well $Q\to \neg P$ is the same as $\neg Q \vee \neg P$ and then
$$\neg (\neg Q \vee \neg P) \equiv Q \wedge P$$
Again, $Q \to (Q \wedge P)$ is the same as $\neg Q \vee (Q \wedge P)$ and this is the same as
$$(\neg Q \vee Q) \wedge (\neg Q \vee P)$$
Now, $\neg Q \vee Q$ is always true and if $P$ is true, also $\neg Q \vee P$ is true.
A: Here's a somewhat automatic way:
\begin{align}
&P \implies (Q \implies \neg (Q \implies \neg P)) \\
&\neg P \vee (\neg Q \vee \neg (\neg Q \vee \neg P)) \\
&\neg P \vee (\neg Q \vee (Q \wedge P)) \\
&\neg P \vee ((\neg Q \vee Q) \wedge (\neg Q \vee P)) \\
&\neg P \vee (1 \wedge (\neg Q \vee P)) \\
&\neg P \vee (\neg Q \vee P) \\
&(\neg P \vee P) \vee \neg Q \\
&1 \vee \neg Q \\
&1
\end{align}
A: 
If $Q$, then statement $\lnot (Q → \lnot P)$ must be true. We know that $P$ is true. For $¬(Q → ¬P)$ to be true, $Q$ must be true. Therefore, given that $P$ is true, statement $Q → ¬(Q → ¬P)$ is true.

No, this does not work as a demonstration. Let's analyze your reasoning:

If $Q$, ...

Your start is ok. That is, given that you need to show $Q \to \neg (Q \to \neg P)$, it makes sense to assume $Q$ is true, and show that in that case $\neg (Q \to \neg P)$ must be true as well.

... then statement $\lnot (Q → \lnot P)$ must be true. ...

No. You need to show that $\lnot (Q → \lnot P)$ must be true.

... We know that $P$ is true. ...

Sure, that is given.

For $¬(Q → ¬P)$ to be true, $Q$ must be true.

True ... but again you are going the wrong way around. You need to show that $\lnot (Q → \lnot P)$ is true ... on the basis of the truth of $Q$ (and the truth of $P$)

Therefore, given that $P$ is true, statement $Q → ¬(Q → ¬P)$ is true.

No, your reasoning hasn't shown this at all.
A: You want to prove $P\land Q$ implies $\neg(Q\to\neg P)$, which is equivalent to $\neg(\neg Q\lor\neg P)$ and hence $P\land Q$.
