How is this disjunctive form found through propositional algebra I'm learning about disjunctive normal form and the algebra of propositions.  The text is Discrete Mathematics with Graph Theory, 3rd Edition by Goodaire and Parmenter (it wasn't highly recommended on Amazon but it is what BSU has chosen).  
In the examples is this implication, $p \rightarrow (q \wedge r)$ which they use to demonstrate how to show in disjunctive normal form using both truth tables and algebra.  I'm lost on the algebra.  Please help me understand where it's coming from.
$
\begin{align}
[p\rightarrow (q\wedge r)] 
&\Leftrightarrow [(\neg p) \vee (q \wedge r)] \\
&\Leftrightarrow [((\neg p)\wedge q)\vee ((\neg p)\wedge (\neg q)) \vee (q \wedge r)] \\
&\Leftrightarrow [((\neg p)\wedge q \wedge r) \vee ((\neg p)\wedge q \wedge (\neg r)) \vee  \\
& \hspace{20pt}((\neg p )\wedge (\neg q) \wedge r) \vee ((\neg p) \wedge (\neg q) \wedge (\neg r)) \vee \\
& \hspace{20pt} (p \wedge q \wedge r) \vee ((\neg p)\wedge q \wedge r)] \\
&\Leftrightarrow [((\neg p)\wedge q \wedge r) \vee ((\neg p)\wedge q \wedge (\neg r)) \vee \\ & \hspace{15pt}((\neg p )\wedge (\neg q) \wedge r) \vee ((\neg p) \wedge (\neg q) \wedge (\neg r)) \vee (p \wedge q \wedge r)]
\end{align}
$
I'm fine up to the first logical equivalance.  However, on the second, I'm lost.  I'm not sure which one of the properties they've discussed was used in this algebra.  Is a distributive property being used?
 A: What is being done, here, in the second  equivalence is using the equivalence of $$\lnot p \equiv \lnot p\land T \equiv (\lnot p \land (\underbrace{q\lor \lnot q}_{\large \text{true}})) \underbrace{\equiv}_{\large \text{DL}} (\lnot p \land q) \lor (\lnot p \land \lnot q)$$
So $$\color{blue}{\lnot p} \lor \color{red}{(q\land r)} \equiv \color{blue}{(\lnot p \land q) \lor (\lnot p \land \lnot q)}\lor \color{red}{(q\land r)}$$
So yes, in a sense, the distributive law (DL) is being used, as well as the tautologies $\lnot p \equiv (\lnot p \land T)$ and $q \lor \lnot q = T$.
The same approach is being used in the second equivalence, which makes use of the tautology $r \lor \lnot r$, etc.
A: He has used the following "tricks" (check them with truth-tables) :

$$p \equiv (p \land T)$$

and 

$$T \equiv (q \lor \lnot q)$$

So, you have the following chain of equivalences :

$$[(\lnot p) \lor (q \land r)] \equiv [(\lnot p \land T) \lor (q \land r)] \equiv [(\lnot p \land (q \lor \lnot q)) \lor (q \land r)] $$

Now, if you "distribute" $\lnot p$, you will get the second equivalence.
