proving logical equivalence $(P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)$ I am currently working through Velleman's book How To Prove It and was asked to prove the following
$(P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)$
This is my work thus far
$(P \to Q) \wedge (Q \to P)$
$(\neg P \vee Q) \wedge (\neg Q \vee P)$ (since $(P \to Q) \equiv (\neg P \vee Q)$)
$\neg[\neg(\neg P \vee Q) \vee \neg (\neg Q \vee P)]$ (Demorgan's Law)
$\neg [(P \wedge \neg Q) \vee (Q \wedge \neg P)]$ (Demorgan's Law)
At this point I am little unsure how to proceed. 
Here are a few things I've tried or considered thus far:
I thought that I could perhaps switch some of the terms in step 3 using the law of associativity however the $\neg$ on the outside of the two terms prevents me from doing so (constructing a truth table for $\neg (\neg P \vee Q) \vee (\neg Q \vee P)$ and $\neg (\neg P \vee \neg Q) \vee \neg (P \vee Q)$ for sanity purposes)
I can't seem to apply the law of distribution since the corresponding terms are negated.
Applying demorgans law to one of the terms individually on step 2 or 3 doesnt seem to get me very far either.
Did I perhaps skip something? Am I even on the right track? Any help is appreciated
 A: 
$$(P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)$$

I'll start with your initial work, but instead of employing DeMorgan's as you did, we'll use the Distributive Law (DL), in two "steps":
$$\begin{align} (P \leftrightarrow Q) &\equiv (P \to Q) \wedge (Q \to P) \tag{correct}\\ \\
&\equiv (\color{blue}{\bf \lnot P \lor Q}) \land (\color{red}{\bf \lnot Q \lor P})\tag{correct} \\ \\
&\equiv \Big[\color{blue}{\bf \lnot P} {\land} \color{red}{\bf(\lnot Q \lor P)}\Big] \color{blue}{\lor} \Big[\color{blue}{\bf Q} \land \color{red}{\bf (\lnot Q \lor P)}\Big]\tag{DL}\\ \\
& \equiv \Big[(\color{blue}{\bf \lnot P} \land \color{red}{\bf\lnot Q)} \lor (\color{blue}{\bf\lnot P} \land \color{red}{\bf P})\Big] \lor \Big[(\color{blue}{\bf Q} \land \color{red}{\bf \lnot Q}) \lor (\color{blue}{\bf Q} \land \color{red}{\bf P})\Big]\tag{DL} \\ \\
&\equiv \Big[({\lnot P}\land \lnot Q) \lor \text{False}\Big] \lor \Big[\text{False} \lor (Q \land P)\Big]\tag{why?} \\ \\
&\equiv (P \land Q) \lor (\lnot P \land \lnot Q)\tag{why?}\end{align}$$
A: That's a great book. You'll learn a lot from it.
Justify the following: $$\begin{align} (\neg P \lor Q) \land (\neg Q \lor P) &\equiv[(\neg P\lor Q) \land \neg Q] \lor [(\neg P\lor Q)\land P] \\
&\equiv [(\neg P \land \neg Q) \lor (Q\land \neg Q)]\lor [(\neg P\land P)\lor (Q\land P)] \\
&\equiv (\neg P\land \neg Q) \lor (Q\land P).\end{align}$$
A: You can also "reverse" amWhy solution, starting with :

$(P \land Q) \lor ( \lnot P \land \lnot Q)$

you can use the distributivity laws to obtain :
$$[(P \land Q) \lor \lnot P] \land [(P \land Q) \lor \lnot Q]$$
and again :
$$(P \lor \lnot P) \land (Q \lor \lnot P) \land (P \lor \lnot Q) \land (Q \lor \lnot Q]$$
i.e.
$$T \land (Q \lor \lnot P) \land (P \lor \lnot Q) \land T$$
i.e.
$$(Q \lor \lnot P) \land (P \lor \lnot Q)$$
Now we use the equivalence between $(\lnot A \lor B) \quad$ and $\quad (A \rightarrow B)$, and get :
$$(P \rightarrow Q) \land (Q \rightarrow P)$$
that is

$P \leftrightarrow Q$.

A: For an alternative approach you can make the truth table of both logical expressions.
$$
\begin{array}{|c|c| c|c| c|c| c|c| c|}
\hline
P                         & Q                        & 
\neg P                    &  \neg Q                  &
(P \wedge  Q)             & (\neg P \wedge \neg Q)   & 
(P \wedge Q ) \vee (\neg P \wedge \neg Q)
\\\hline
 V                         & V                    
&F                         & F                   
&V                         & F                   
&V
\\\hline
V                         & F                   & 
F                         & V                   &
F                        & F                   & 
F
\\\hline
F                         & V                   & 
V                         & F                   &
F                         & F                   &
F 
\\\hline
F                         & F                   & 
V                         & V                   &
F                         & V                   & 
V
\\\hline
\end{array}
$$
and
$$
\begin{array}{|c|c|c|}
\hline
 P&Q& P \leftrightarrow   Q 
\\\hline
 V&V&V
\\\hline
 V&F&F 
\\\hline
 F&V&F
\\\hline
 F&F&V
\\\hline
\end{array}
$$
