# Existence of Conjunctive Normal Form of Statement

## Theorem

Any propositional formula can be expressed in conjunctive normal form (CNF).

## Proof

A propositional variable is already trivially in conjunctive normal form (CNF).

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:

$P_1 \land P_2 \land \cdots \land P_n$

where $P_1, P_2, \ldots, P_n$ are either:

• Literals;
• Statements of the form $\left({Q_1 \lor Q_2 \lor \ldots \lor Q_n}\right)$

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

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

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

$Q_1 \lor Q_2 \lor \ldots \lor \left({R_1 \land R_2 \land \ldots \land R_m}\right) \ldots \lor Q_n$

into:

 $\ds$  $\ds \left({Q_1 \lor Q_2 \lor \ldots \lor Q_n \lor R_1}\right)$ $\ds$ $\land$ $\ds \left({Q_1 \lor Q_2 \lor \ldots \lor Q_n \lor R_2}\right)$ $\ds$ $\land$ $\ds \ldots$ $\ds$ $\land$ $\ds \left({Q_1 \lor Q_2 \lor \ldots \lor Q_n \lor R_m}\right)$

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

It can be seen then that each of the

$\left({Q_1 \lor Q_2 \lor \ldots \lor Q_n \lor R_k}\right)$

are terms in the CNF expression required.

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

$\blacksquare$