Why is "$P \Rightarrow Q$" equivalent to "$\neg Q \Rightarrow \neg P$"? Why is "$P \Rightarrow Q$" equivalent to "$\neg Q \Rightarrow \neg P$"?
I am trying to understand why this does always apply, in terms of pure logic. Can you please explain it to me?
 A: Suppose it's true that:


*

*If you hit my glass table with a hammer, it will break.

*My glass coffee table is not broken.


If you accept premisses 1 and 2, wouldn't you conclude from this that
3. My glass coffee table has not been hit with a hammer?

If we let $P$ represent “my glass table was hit with a hammer” and $Q$ be “my glass table is broken”, then (1) is $P\implies Q$, and (2) is $\lnot Q$.
From (1) and (2) we can conclude that the table has not been hit with a hammer; that's $\lnot P$.  One way to write this is $$((P\implies Q) \land \lnot Q) \implies \lnot P$$
but is $$(P\implies Q)\implies (\lnot Q \implies \lnot P)$$ which says that if we know $P\implies Q$ (that's (1)) then if we know $\lnot Q$ (that's (2)) then we can conclude $\lnot P$ (that's (3)).
On the other hand, it is not the case that 
$$(P\implies Q)\implies (\lnot P \implies \lnot Q)$$
This says that if (1) is true, and you know my table was not hit with a hammer, then you know it is not broken.  But no, that's wrong, because actually it is broken because my brother-in-law got drunk and fell on it; no hammer was involved.
A: It is given that $P\to Q$. By material implication, $\neg P\vee Q$. By commutativity, $Q\vee\neg P$. Again, by material implication, $\neg Q\to \neg P$.
A: $$\begin{array} {|c|c|c|c|} \hline
P & Q & P \Rightarrow Q & \lnot Q \Rightarrow \lnot P \\ \hline
\top & \top & \top & \top \\
\top & \bot & \bot & \bot \\
\bot & \top & \top & \top \\
\bot & \bot & \top & \top \\
\hline
\end{array}$$
Equal.
A: You can check it simply by writing up every situation in a truth-table and see that they produce the same results for any choices of $ P $ and $ Q $. 
$ P \Rightarrow Q $ is only false when $ P $ is true and $ Q $ is false. 
Likewise $ \neg Q \Rightarrow \neg P $ is only false when $ \neg Q $ is true and $ \neg P $ is false, thus when $ Q $ is false and $ P $ is true.
