Does (If not P then Q) imply (If P then Q)? My truth table says yes but I want verification As the title says, is this true? 
$$(\lnot P \to \lnot Q) \to (P \to Q)$$
The truth table is
\begin{array}{rrrrrr}
              P & Q & \lnot P & \lnot Q & \lnot P \to \lnot Q & P \to Q & (\lnot P \to \lnot Q) \to (P \to Q)   \\ \hline
           T &   T &   F &   F &   T &   T &  T \\
           T &   F &   F &   T &   T &   F &  F \\
           F &   T &   T &   F &   F &   T &  T \\
           F &   F &   T &   T &   T &   T & T \\ 
\end{array}
It seems like it's true from the table.
If it is true, is it true because $$(\lnot P \to \lnot Q) \to (P \to Q)$$ has the same truth table corresponding to the $\to$ connective which is false only when the antecedent is T but the consequent is F?
Or is it true because the statement is true when the premises of $\lnot P \to \lnot Q$ and $P \to Q$ are true?
If it's not true, why not?
 A: $(\lnot P\to\lnot Q)\to(P\to Q)$ is not a tautology because it is not true when $P$ is true but $Q$ is false.   That is shown in the second row of your truth table.
Likewise, it is not a contradiction.   The statement is conditionally true.
The statement is logically equivalent to $\lnot(P\land\lnot Q)$, also to $(\lnot P\lor Q)$.

Now $(\lnot P\to\lnot Q)\to(Q\to P)$ is a tautology in classical logic.   Notice the order of the terms.
Indeed $\lnot P\to \lnot Q$ is the contrapositive of $Q\to P$, and the two are logically equivalent.
A: No, if we have a statement "$P$ then $Q$", then "$\neg P$ then $\neg Q$" is the inverse of the statement. The inverse being true does not imply the statement is true. 
For instance consider a class where the cutoff for an $A$ is $90\%$. Consider the statement $$
\text{"If you have above an }80\%\text{, then you will receive an }A\text{."}
$$ 
This statement is not true. However its inverse is true.
$$
\text{"If you do not have above an }80\%\text{, then you will not receive an }A\text{."}
$$ 
