How to prove statement: If A, then if B, then C. I was wondering. In general, what are the approaches to proving nested if statements when doing a direct proof?
Thanks.
 A: For "nested if statements", In general:
\begin{align}
P_1\to(P_2\to\dots (P_{n-1}\to(P_n\to Q))))
&\Leftrightarrow (P_1\land P_2\land \dots \land P_{n-1}\land P_n)\to Q\tag*{$(1)$}\\
&\Leftrightarrow \neg P_1\lor\neg P_2\lor \dots \lor\neg P_{n-1}\lor\neg P_n\lor Q\tag*{$(2)$}\\
&\not\Leftrightarrow P_1\land P_2\land \dots \land P_{n-1}\land P_n\land\neg Q\tag*{$(3)$}\\
\end{align}
For $(1)$ it's a direct proof, basicly assume $P_1\dots P_n$, then prove $Q$ is a logical consequence of them.
$(2)$ can also be written as:
$$\neg Q\to(\neg P_1\lor\neg P_2\lor \dots \lor\neg P_{n-1}\lor\neg P_n)$$
In this case, assume $\neg Q$, if this implies any one in $\neg P_1\dots\neg P_n$ hold, it proves the statements.
Sometimes if it's hard to construct a direct proof, $(3)$ would also be a good choice, which is the negation of $(1)$, first assume that $P_1\dots P_n$, and $\neg Q$. If you can prove it's false, then that implies $(1)$ is true.
A: Assuming the 'inside' conditional is the consequent of the 'outside' conditional, you can do a conditional proof inside a conditional proof. Formally, it would look like:

A: Note that $a \rightarrow (b \rightarrow c) \Leftrightarrow (a \wedge b) \rightarrow c$ by the exportation law. The following are common ways to prove such a statement:
Direct Proof
Assume $a$ and $b$, and then derive $c$. 
Proof by Contrapositive
Prove the contrapositive of $(a \wedge b) \rightarrow c$, which is $\neg c \rightarrow \neg(a \wedge b) \Leftrightarrow \neg c \rightarrow (\neg a \vee \neg b)$. You can do this by assuming $\neg c$ and then deriving $\neg a \vee \neg b$.
Proof by Contradiction
Assume the negation of $(a \wedge b) \rightarrow c$, which is $a \wedge b \wedge \neg c$, and then derive a contradiction.
