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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.[citation needed] 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.[1]: 153  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:

  1. DNF → (Conjunction) DNF
  2. DNF → (Conjunction)
  3. ConjunctionLiteral Conjunction
  4. ConjunctionLiteral
  5. LiteralVariable
  6. LiteralVariable

Where Variable is any variable.

For example, all of the following formulas are in DNF:

However, the following formulas are not in DNF:

The formula is in DNF, but not in full DNF; an equivalent full-DNF version is .

Conversion to DNF

Karnaugh map of the disjunctive normal form A∧¬B∧¬D)ABC)(ABD)(A∧¬B∧¬C)
Karnaugh map of the disjunctive normal form AC∧¬D)(BCD)(A∧¬CD)B∧¬C∧¬D). Despite the different grouping, the same fields contain a "1" as in the previous map.

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.[1]: 152–153  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[note 1] 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.

Computational complexity

The Boolean satisfiability problem on conjunctive normal form formulas is NP-hard; by the duality principle, so is the falsifiability problem on DNF formulas. Therefore, it is co-NP-hard to decide if a DNF formula is a tautology.

Conversely, a DNF formula is satisfiable if, and only if, one of its conjunctions is satisfiable; this can be decided in polynomial time.[2]


An important variation used in the study of computational complexity is k-DNF. A formula is in k-DNF if it is in DNF and each conjunction contains at most k literals.

See also


  1. ^ Ignoring variations based on associativity and commutativity of AND and OR.


  1. ^ a b B.A. Davey and H.A. Priestley (1990). Introduction to Lattices and Order. Cambridge Mathematical Textbooks. Cambridge University Press.
  2. ^ Martin Zimmermann (2015-01-22). "The Complexity of Counting Models of Linear-time Temporal Logic" (PDF). Saarland University. Retrieved 2023-02-02.