Reverse contraposition: how to prove $(\neg Q \implies \neg P)\vdash(P \implies Q)$? Contraposition states $(P \implies Q)\vdash(\neg Q \implies \neg P)$. But how to prove when premise and conclusion are reversed, namely how to prove $(\neg Q \implies \neg P)\vdash(P \implies Q)$?
I have nice example of this:
If rains, then the floor is wet. And, of course, if floor is not wet, then it doesn't rain. (Observe that reverse statement is not true: if floor is wet, then it rains is a false statement.)
If the floor is not wet, then it doesn't rain. (Again, observe that reverse statement is not true: if it doesn't rain, then the floor is not wet is a false statement).
 A: One way to prove this is by using a proof tree. You start with the negation of your formula then apply some contradiction-hunting rules to show that that negation cannot be true; here:

Here is where the tree was generated.
A: I found simple heuristic argument for my own question. This is of course, just an intuition not the 'real' proof, but I believe that such basic results must have common-sense explanations for all self-learners, as am I.
Notation $A_1, A_2, ... \implies B$ says that if some $A_i$'s hold then $B$ must also hold; therefore the consequence of every $A_i$ is $B$ and other way around, causes of $B$ are all $A_i$'s.
Now, what about relations between $\neg B$ and $A_i$'s?  Other way to say $\neg B$ is that $B$ doesn't hold which is same as saying that consequence of some $A_i$'s doesn't hold. This implies that $A_i$ doesn't hold either because if some $A_i$'s do indeed hold we would get $B, \neg B$ at the both holds at the same time, which is not possible.
To prove my question, as said in comments, we can just set $\neg Q := P, \neg P := Q$ and using $\neg \neg A = A$ we are done.
