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.[1] As a normal form, it is useful in automated theorem proving.

## Definition

A logical formula is considered to be in DNF if it is a disjunction of one or more conjunctions of one or more literals.[2][3][4] A DNF formula is in full disjunctive normal form if each of its variables appears exactly once in every conjunction and each conjunction appears at most once (up to the order of variables). As in conjunctive normal form (CNF), the only propositional operators in DNF are and (${\displaystyle \wedge }$), or (${\displaystyle \vee }$), and not (${\displaystyle \neg }$). 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) ${\displaystyle \vee }$ DNF
2. DNF → (Conjunction)
3. ConjunctionLiteral ${\displaystyle \wedge }$ Conjunction
4. ConjunctionLiteral
5. Literal${\displaystyle \neg }$Variable
6. LiteralVariable

Where Variable is any variable.

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

• ${\displaystyle (A\land \neg B\land \neg C)\lor (\neg D\land E\land F\land D\land F)}$
• ${\displaystyle (A\land B)\lor (C)}$
• ${\displaystyle (A\land B)}$
• ${\displaystyle (A)}$

The formula ${\displaystyle A\lor B}$ is in DNF, but not in full DNF; an equivalent full-DNF version is ${\displaystyle (A\land B)\lor (A\land \lnot B)\lor (\lnot A\land B)}$.

The following formulas are not in DNF:[5]

• ${\displaystyle \neg (A\lor B)}$, since an OR is nested within a NOT
• ${\displaystyle \neg (A\land B)\lor C}$, since an AND is nested within a NOT
• ${\displaystyle A\lor (B\land (C\lor D))}$, since an OR is nested within an AND

## Conversion to DNF

In classical logic each propositional formula can be converted to DNF[6] ...

### ... by syntactic means

The conversion involves using logical equivalences, such as double negation elimination, De Morgan's laws, and the distributive law. Formulas built from the primitive connectives ${\displaystyle \{\land ,\lor ,\lnot \))$[7] can be converted to DNF by the following canonical term rewriting system:[8]

## Maximum number of conjunctions

Any propositional formula is built from ${\displaystyle n}$ variables, where ${\displaystyle n\geq 1}$.

There are ${\displaystyle 2n}$ possible literals: ${\displaystyle L=\{p_{1},\lnot p_{1},p_{2},\lnot p_{2},\ldots ,p_{n},\lnot p_{n}\))$.

${\displaystyle L}$ has ${\displaystyle (2^{2n}-1)}$ non-empty subsets.[18]

This is the maximum number of conjunctions a DNF can have.[12]

A full DNF can have up to ${\displaystyle 2^{n))$ conjunctions, one for each row of the truth table.

Example 1

Consider a formula with two variables ${\displaystyle p}$ and ${\displaystyle q}$.

The longest possible DNF has ${\displaystyle 2^{(2\times 2)}-1=15}$ conjunctions:[12]

${\displaystyle {\begin{array}{lcl}(\lnot p)\lor (p)\lor (\lnot q)\lor (q)\lor \$$\lnot p\land p)\lor {\underline {(\lnot p\land \lnot q)))\lor {\underline {(\lnot p\land q)))\lor {\underline {(p\land \lnot q)))\lor {\underline {(p\land q)))\lor (\lnot q\land q)\lor \\(\lnot p\land p\land \lnot q)\lor (\lnot p\land p\land q)\lor (\lnot p\land \lnot q\land q)\lor (p\land \lnot q\land q)\lor \\(\lnot p\land p\land \lnot q\land q)\end{array))}$ The longest possible full DNF has 4 conjunctions: they are underlined. This formula is a tautology. Example 2 Each DNF of the e.g. formula ${\displaystyle (X_{1}\lor Y_{1})\land (X_{2}\lor Y_{2})\land \dots \land (X_{n}\lor Y_{n})}$ has ${\displaystyle 2^{n))$ conjunctions. ## Computational complexity The Boolean satisfiability problem on conjunctive normal form formulas is NP-complete. 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 simply by checking that at least one conjunction does not contain conflicting literals. ## Variants 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.[19] ## See also ## Notes 1. ^ 2. ^ Davey & Priestley 1990, p. 153. 3. ^ Gries & Schneider 1993, p. 67. 4. ^ Whitesitt 2012, pp. 33–37. 5. ^ However, they are in negation normal form. 6. ^ Davey & Priestley 1990, p. 152-153. 7. ^ Formulas with other connectives can be brought into negation normal form first. 8. ^ Dershowitz & Jouannaud 1990, p. 270, Sect.5.1. 9. ^ 10. ^ ${\displaystyle \phi }$ = ((NOT (p AND q)) IFF ((NOT r) NAND (p XOR q))) 11. ^ like ${\displaystyle (a\land b)\lor (b\land a)\lor (a\land b\land b)}$ 12. ^ a b c It is assumed that repetitions and variations[11] based on the commutativity and associativity of ${\displaystyle \lor }$ and ${\displaystyle \land }$ do not occur. 13. ^ a b Halbeisen, Lorenz; Kraph, Regula (2020). Gödel´s theorems and zermelo´s axioms: a firm foundation of mathematics. Cham: Birkhäuser. p. 27. ISBN 978-3-030-52279-7. 14. Howson, Colin (1997). Logic with trees: an introduction to symbolic logic. London ; New York: Routledge. p. 41. ISBN 978-0-415-13342-5. 15. ^ a b Cenzer, Douglas; Larson, Jean; Porter, Christopher; Zapletal, Jindřich (2020). Set theory and foundations of mathematics: an introduction to mathematical logic. New Jersey: World Scientific. pp. 19–21. ISBN 978-981-12-0192-9. 16. ^ a b Halvorson, Hans (2020). How logic works: a user's guide. Princeton Oxford: Princeton University Press. p. 195. ISBN 978-0-691-18222-3. 17. ^ That is, the language with the propositional variables ${\displaystyle A,B,C,\ldots }$ and the connectives ${\displaystyle \{\land ,\lor ,\neg$$)$.
18. ^ ${\displaystyle \left|{\mathcal {P))(L)\right|=2^{2n))$
19. ^