Existence of Disjunctive Normal Form of Statement

Theorem
Any propositional formula can be expressed in disjunctive normal form (DNF).

Proof
A propositional variable is already trivially in disjunctive normal form (DNF).

So we consider the general propositional formula $S$.

First we convert to negation normal form (NNF).

This is always possible, by Existence of Negation Normal Form of Statement.

Now $S$ will be of the form:
 * $P_1 \lor P_2 \lor \cdots \lor P_n$

where $P_1, P_2, \ldots, P_n$ are either:
 * Literals;
 * Statements of the form $\left({Q_1 \land Q_2 \land \ldots \land Q_n}\right)$

If all the $Q_1, \ldots, Q_n$ are literals we have finished.

Otherwise they will be of the form $Q_j = \left({R_1 \lor R_2 \lor \ldots \lor R_m}\right)$

If the latter is the case, then use Conjunction Distributes over Disjunction to convert:
 * $Q_1 \land Q_2 \land \ldots \land \left({R_1 \lor R_2 \lor \ldots \lor R_m}\right) \ldots \land Q_n$

into:

It is taken for granted that Conjunction is Associative and Conjunction is Commutative.

It can be seen then that each of the
 * $\left({Q_1 \land Q_2 \land \ldots \land Q_n \land R_k}\right)$

are terms in the DNF expression required.

If any terms are still not in the correct format, then use the above operation until they are.