What does it mean to "show algebraically" in propositional logic?
Although you haven't posted any of the logical rules the problem statement refers to, it is reasonable to assume they are similar to those found in many standard logic textbooks. To prove your identity, I will make use of
$$
\begin{align}
P \land Q &amp; \equiv Q \land P, &amp;\text{$\land$ commutativity} \\
P \land (Q \lor R) &amp; \equiv (P \land Q) \lor (P \land R),  &amp;\text{$\land$ distribution} \\
P \land \lnot P &amp; \equiv \bot  &amp;\text {(3)} \\
P \lor \bot &amp;\equiv P, &amp; \text{$\bot$ annihilation}
\end{align}
$$
where $\bot$ denotes falsehood. With that in mind
$$
\begin{align}
p \iff q \equiv &amp; \\
(\lnot p \lor q) \land (\lnot q \lor p) \equiv &amp;  &amp; \text{By definition}\\
((\lnot p \lor q) \land \lnot q) \lor ((\lnot p \lor q) \land p) \equiv &amp; &amp;\text{By distribution of $\land$} \\
(\lnot q \land (\lnot p \lor q)) \lor (p \land (\lnot p \lor q)) \equiv &amp; &amp;\text{By commutativity of $\land$} \\
(\lnot q \land \lnot p \lor \lnot q \land q) \lor (p \land \lnot p \lor p \land q) \equiv &amp; &amp;\text{By distribution of $\land$} \\
(\lnot q \land \lnot p \lor \bot) \lor (\bot \lor p \land q) \equiv &amp; &amp;\text{By $(3)$} \\
(\lnot q \land \lnot p) \lor (p \land q) \equiv &amp; &amp;\text{By $\bot$ annihilation} \\
(\lnot p \land \lnot q) \lor (q \land p) \equiv &amp; &amp;\text{By commutativity of $\land$} \\
&amp; &amp;\square
\end{align}
$$
I've assumed throughout $\land$ to have higher precedence than $\lor$.