Disjunctive normal form

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 (${\displaystyle \wedge }$), 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. Conjunction → Literal ${\displaystyle \wedge }$ Conjunction
4. Conjunction → Literal
5. Literal → ${\displaystyle \neg }$Variable
6. Literal → Variable

Where Variable is any variable.

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

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

However, the following formulas are not in DNF:

• ${\displaystyle \neg (A\lor B)}$, 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

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

Karnaugh map of the disjunctive normal form (¬A∧¬B∧¬D) ∨ (¬A∧B∧C) ∨ (A∧B∧D) ∨ (A∧¬B∧¬C)

Karnaugh map of the disjunctive normal form (¬A∧C∧¬D) ∨ (B∧C∧D) ∨ (A∧¬C∧D) ∨ (¬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 ${\displaystyle (X_{1}\lor Y_{1})\land (X_{2}\lor Y_{2})\land \dots \land (X_{n}\lor Y_{n})}$ 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.

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.

• Algebraic normal form – an XOR of AND clauses
• Blake canonical form – DNF including all prime implicants
  • Quine–McCluskey algorithm – algorithm for calculating prime implicants
• Propositional logic
• Truth table

• David Hilbert; Wilhelm Ackermann (1999). Principles of Mathematical Logic. American Mathematical Soc. ISBN 978-0-8218-2024-7.
• J. Eldon Whitesitt (24 May 2012). Boolean Algebra and Its Applications. Courier Corporation. ISBN 978-0-486-15816-7.
• Colin Howson (11 October 2005). Logic with Trees: An Introduction to Symbolic Logic. Routledge. ISBN 978-1-134-78550-6.
• David Gries; Fred B. Schneider (22 October 1993). A Logical Approach to Discrete Math. Springer Science & Business Media. pp. 67–. ISBN 978-0-387-94115-8.

• "Disjunctive normal form", Encyclopedia of Mathematics, EMS Press, 2001 [1994]