Existence of Negation Normal Form of Statement

Theorem
Any statement form can be expressed in negation normal form (NNF).

Proof
A simple statement is already trivially in negation normal form (NNF).

So we consider the general compound statement $S$.

In the following, $P$ and $Q$ are to stand for general statement forms, simple or compound.

First, from Functionally Complete Logical Connectives: Conjunction, Negation and Disjunction we have that:
 * $\left\{{\neg, \land, \lor}\right\}$

forms a functionally complete set of logical connectives.

So we can convert $S$ so that:
 * By the definition of the biconditional, instances of $P \iff Q$ can be converted into $\left({P \implies Q}\right) \land \left({Q \implies P}\right)$;
 * By the Rule of Material Implication, instances of $P \implies Q$ can be converted into $\neg P \lor Q$.

Other connectives can likewise be treated appropriately.

Thus $S$ will contain nothing but connectives from $\left\{{\neg, \land, \lor}\right\}$.

Next we can replace:
 * Instances of $\neg \left({P \land Q}\right)$ with $\neg P \lor \neg Q$, by De Morgan's Laws: Disjunction of Negations;
 * Instances of $\neg \left({P \lor Q}\right)$ with $\neg P \land \neg Q$, by De Morgan's Laws: Conjunction of Negations;
 * Instances of $\neg \neg P$ with $P$ by Double Negation Elimination.

At any stage where a negation appears before a parenthesis, it will then appear before the statements inside the parenthesis.

Thus the negation signs gradually move inwards or are eliminated.

Eventually all remaining negation signs will appear next to simple statements.

Hence the result.