alternative rule for negation introduction I have the standard rule for negation introduction, namely:
$$\frac{P\Rightarrow Q\quad P\Rightarrow\neg Q}{\neg P}\quad\text{[Proof by negation]}$$
Now I need a slightly different rule (I'm not sure whether you'd say it's stronger or weaker):
$$\frac{P\Rightarrow Q\quad\neg Q}{\neg P}\quad\text{[Modus tollens]}$$
Can I derive the former from the latter? I'm guessing you can do it if you assume something like the law of the excluded middle. If you can get $P\Rightarrow\neg Q$ of course then you'd be done.
 A: Together with the proof given in @Hailey's answer, it only remains to show that $\neg P\vee Q$ can be derived from $P\Rightarrow Q$. For this you need the law of the excluded middle.
For the proof we have the premise $P\Rightarrow Q$. Now suppose $P$ holds, then by modus ponens and the premise we have that $Q$ holds. From this $\neg P\vee Q$ holds by conjunction introduction. Therefore $P\Rightarrow\neg P\vee Q$. Now also $\neg P\Rightarrow\neg P$. Therefore $\neg P\Rightarrow\neg P\vee Q$ again by conjunction introduction. From these two implications $P\vee\neg P\Rightarrow\neg P\vee Q$ can be derived, which is a form of disjunction introduction which I leave out here. Lastly by the law of the excluded middle we can take $P\vee\neg P$ and together with modus ponens we have $\neg P\vee Q$, as required.
[Note: This has been voted up, however the fully correct answer is the one that begins 'The trick is...'.]
A: The trick is to realise that $B\Rightarrow(A\Rightarrow B)$ is a tautology. To see this:
Suppose
  B
Hence
  Suppose
    A
  Hence
    B // because B holds by the outer supposition
  A=>B
B=>(A=>B)

I'm not entirely happy with this, because it seems that you are not deriving that B holds by applying any rule whose premises include A, you are simply restating the outer supposition. Apparently though this is fine. As Von Neumann once said, in mathematics you don't understand things, you just get used to them.
Now the result follows directly from modus ponens. Formally:
$$
\frac{B\Rightarrow(A\Rightarrow B)\quad B}{A\Rightarrow B}
$$
Again, since the left hand premise is a tautology you can leave it out, so you get:
$$
\frac{B}{A\Rightarrow B}
$$
This means that $\neg Q$ can be replaced with $P\Rightarrow\neg Q$ in the question, in which case the rule just becomes the standard one for proof by negation.
A: $P\implies Q$  is equivalent to $\neg P  ∨Q $
So, $(\neg P  ∨ Q )∧\neg Q=(\neg P∧\neg Q )∨(\neg Q ∧ Q)=(\neg P∧\neg Q )\quad$  [As $(Q∧\neg Q)=false$]
Now, $\neg Q $ is given that means,it implies $\neg P$
