Converting $p \leftrightarrow q$ to disjunctive normal form

Question

I started like this: \begin{align*} p &\leftrightarrow q\\ (p \to q) &\wedge (q \to p)\\ (q \vee \neg p) &\wedge (p \vee \neg q) \end{align*} But I'm not sure how to continue and how to change this formula to disjunctive normal form.

Answer 1

To get to DNF: \begin{align*} &p \leftrightarrow q\\ \equiv &(p \to q) \wedge (q \to p) \qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\,\text{(Definition of iff)}\\ \equiv &(\neg p \vee q) \wedge (\neg q \vee p) \qquad\qquad\qquad\qquad\qquad\qquad\,\,\text{(Eliminate $\to$)}\\ \equiv &((\neg p \wedge \neg q)\vee (\neg p\wedge p)) \vee ((q \wedge \neg q)\vee (q \wedge p)) \qquad\,\,\,\,\,\,\, \text{(Distribution)}\\ \equiv &((\neg p \wedge \neg q)\vee \textrm{F}) \vee (\textrm{F}\vee (q \wedge p)) \qquad\text{(Since $(\neg p \wedge p)=(q \wedge \neg q)=\textrm{F}$))}\\ \equiv &(\neg p \wedge \neg q) \vee (q \wedge p) \qquad\qquad \text{(Since $(\neg p \wedge \neg q)\vee \textrm{F}=(\neg p \wedge \neg q)$, and $\textrm{F}\vee (q \wedge p)=(q \wedge p))$} \end{align*}