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

or:

statements of the form $\paren {Q_1 \land Q_2 \land \ldots \land Q_n}$

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

Otherwise they will be of the form $Q_j = \paren {R_1 \lor R_2 \lor \ldots \lor R_m}$

If the latter is the case, then use Conjunction Distributes over Disjunction to convert:

$Q_1 \land Q_2 \land \ldots \land \paren {R_1 \lor R_2 \lor \ldots \lor R_m} \ldots \land Q_n$

into:

 $\ds$  $\ds \paren {Q_1 \land Q_2 \land \ldots \land Q_n \land R_1}$ $\ds$ $\lor$ $\ds \paren {Q_1 \land Q_2 \land \ldots \land Q_n \land R_2}$ $\ds$ $\lor$ $\ds \ldots$ $\ds$ $\lor$ $\ds \paren {Q_1 \land Q_2 \land \ldots \land Q_n \land R_m}$

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

It can be seen then that each of the

$\paren {Q_1 \land Q_2 \land \ldots \land Q_n \land R_k}$

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.

$\blacksquare$