Summary
In boolean logic, a disjunctive normal form (DNF) is a canonical normal form of a logical formula consisting of a disjunction of conjunctions; it can also be described as an OR of ANDs, a sum of products, or (in philosophical logic) a cluster concept. As a normal form, it is useful in automated theorem proving. A logical formula is considered to be in DNF if it is a disjunction of one or more conjunctions of one or more literals. A DNF formula is in full disjunctive normal form if each of its variables appears exactly once in every conjunction. As in conjunctive normal form (CNF), the only propositional operators in DNF are and (), or (), and not (). The not operator can only be used as part of a literal, which means that it can only precede a propositional variable. The following is a context-free grammar for DNF: DNF → (Conjunction) DNF DNF → (Conjunction) Conjunction → Literal Conjunction Conjunction → Literal Literal → Variable Literal → Variable Where Variable is any variable. For example, all of the following formulas are in DNF: However, the following formulas are not in DNF: since an OR is nested within a NOT since an AND is nested within a NOT since an OR is nested within an AND The formula is in DNF, but not in full DNF; an equivalent full-DNF version is . Converting a formula to DNF involves using logical equivalences, such as double negation elimination, De Morgan's laws, and the distributive law. All logical formulas can be converted into an equivalent disjunctive normal form. However, in some cases conversion to DNF can lead to an exponential explosion of the formula. For example, converting the formula to DNF yields a formula with 2n terms. Every particular Boolean function can be represented by one and only one full disjunctive normal form, one of the canonical forms. In contrast, two different plain disjunctive normal forms may denote the same Boolean function; see the illustrations. The Boolean satisfiability problem on conjunctive normal form formulas is NP-hard; by the duality principle, so is the falsifiability problem on DNF formulas.
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