How do we know that the contrapositive, ¬q → ¬p, of a conditional statement p → q always has the same truth value as p → q? 
I am having a bit of trouble understanding the pasted excerpt. I think I might be missing something basic. As I understand it, the contrapositive of a conditional statement is where we take a conditional statement and both 1) flip the hypothesis and conclusion and 2) negate the q and p so we have  ¬q -> ¬p
Looking at the truth table of the original p -> q I can convert each possibility to the contrapositive ¬q -> ¬p . So, for example, when p is True and q is False, the p -> q is false. I can now turn this case into the contrapositive by taking the q and negating it which is True and then take the p and negating it which is False.
What does this mean though that the contrapositive has the same truth value as p -> q? Like, the truth of table of p -> q was just a fact that was given to me. How do I know what the truth value for each possibility of ¬q -> ¬p is though? Is it simply that we can always convert the contrapositive back to the p -> q statement by "un-negating" the q and p in ¬q -> ¬p and then know the truth value based on original truth table for p -> q where we know it's only False when p is True and q is False?
Just generally confused. I hope my rambling question makes some sense.
Thanks in advance.
 A: Well, you can easily verify that the two statements are equivalent using a truth table, as you have done.
You might also use the definition $$p\rightarrow q \equiv \sim p \cup q$$ to show that both are equivalent.
If you want to gain some intuitive sense, $p\rightarrow q$ means "if $p$, then $q$". This means that the truth of $p$ indicates the truth of $q$. Whereas the second statement means "if not $q$, then not $p$". The two are clearly equivalent. To see this take, as an example, $p\equiv$ "it is cloudy" and $q\equiv$ "it is raining".
A: I think something fundamental may be slipping past you.  As I think you understand, the implication $p \Rightarrow q$ means that either the premise (which we're calling $p$) is false or the consequence (what we're calling $q$) is true (or possibly both).
But with this understanding, the truth values of the contrapositive are not arbitrary.  Let's look at the contrapositive $\lnot q \Rightarrow \lnot p$.  By our definition of an implication, this means the premise (whatever it may be) is false or the consequence (whatever it may be) is true.  In this case, the premise is $\lnot q$, and that's false exactly when $q$ is true.  Similarly, the consequence of the contrapositive is $\lnot p$, and that's true exactly when $p$ itself is false.
That means that $\lnot (\lnot q) \lor \lnot p$ (which is equivalent to the contrapositive) is true exactly when $\lnot p \lor q$ is true, so the contrapositive is true exactly when the original implication is true.
