Proving equivalence of statements without using a truth table How can I prove these statements without using a truth table?
$1.\quad (p \leftrightarrow q) \equiv (p \land q) \lor (\neg p \land \neg q)$ 
$2. \quad (p \land q \land r) \equiv \neg p \lor \neg q \lor \neg r$
$3.\quad p \land (p \lor r) \equiv p$
 A: You've posted three separate questions in one question field. 
So I will not answer them all.  I will give hints for each:
$(1)\;$ Note that by definition, 
$$\begin{align} p\leftrightarrow q &\equiv (p\to q) \land (q \to p)\tag{Definition of biconditional}\\ \\
&\equiv (\lnot p \lor q) \land (\lnot q \lor p)\tag{implication}
\end{align}$$
Use the distributive property (a couple of times, if needed) we can work to show it is  equivalent to  $$(p\land q) \lor (\lnot p \land \lnot q)$$   I'll let you take it from here.

$(2)\;$ Use DeMorgan's twice (and make use of the associative property), on $\neg (p \land q \land r),\,$ to show that is equivalent to  $\;\neg p \lor \neg q \lor \neg r.$  
$$\begin{align}
\neg (p \land q \land r)
&\equiv \lnot ((p\land q) \land r)\tag{associativity}\\ \\ 
&\equiv (\lnot (p\land q) \lor \lnot r) \tag{DeMorgan's}\\ \\ 
&\equiv (\lnot p \lor \lnot q)\lor \lnot r \tag{DeMorgan's again}\\ \\
&\equiv \lnot p \lor \lnot q \lor \lnot r\tag{associativity}
\end{align}$$

$(3)\;$ use the distributive property  on $\;p \land (p \lor r)\;$ to show it is equivalent to $\;p.$ 
$$\begin{align} 
p \land (p \lor r)&\equiv (p\land p) \lor (p \land r)\tag{distributive property}\\ \\  &\equiv p \lor (p\land r) \tag{Why is $p\land p \equiv p$?}
\end{align}$$
I'll let you finish the proof for $(3)$.
