Inboolean logic, adisjunctive normal form (DNF) is anormal form of a logical formula consisting of a disjunction of conjunctions; it can also be described as anOR of ANDs, asum of products, or — inphilosophical logic — acluster concept.^[1] The disjunctive normal form and its counterpart, theconjunctive normal form, are the most common standardized ways of representingboolean expressions. They are widely used in various applications such ascircuit design orautomated theorem proving.

A logical formula is considered to be in DNF if it is adisjunction of one or moreconjunctions of one or moreliterals.^[2][3][4] A DNF formula is infull 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 inconjunctive normal form (CNF), the only propositional operators in DNF areand (${\displaystyle \wedge }$),or (${\displaystyle \vee }$), andnot (${\displaystyle \mathrm{\neg }}$). Thenot operator can only be used as part of a literal, which means that it can only precede apropositional variable.

The following is acontext-free grammar for DNF:

DNF${\displaystyle \to }$ (Disjunct)${\displaystyle \mid }$ (Disjunct)${\displaystyle \vee }$DNF

Disjunct${\displaystyle \to }$Literal${\displaystyle \mid }$Literal${\displaystyle \wedge }$Disjunct

Literal${\displaystyle \to }$Variable${\displaystyle \mid }$${\displaystyle \mathrm{\neg }}$Variable

WhereVariable is any variable.

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

• ${\displaystyle (A\wedge \mathrm{\neg }B\wedge \mathrm{\neg }C)\vee (\mathrm{\neg }D\wedge E\wedge F\wedge D\wedge F)}$
• ${\displaystyle (A\wedge B)\vee (C)}$
• ${\displaystyle (A\wedge B)}$
• ${\displaystyle (A)}$

The formula${\displaystyle A\vee B}$ is in DNF, but not in full DNF; an equivalent full-DNF version is${\displaystyle (A\wedge B)\vee (A\wedge \mathrm{\neg }B)\vee (\mathrm{\neg }A\wedge B)}$.

The following formulas arenot in DNF:

• ${\displaystyle \mathrm{\neg }(A\vee B)}$, since an OR is nested within a NOT
• ${\displaystyle \mathrm{\neg }(A\wedge B)\vee C}$, since an AND is nested within a NOT
• ${\displaystyle A\vee (B\wedge (C\vee D))}$, since an OR is nested within an AND^[5]

Inclassical logic each propositional formula can be converted to DNF^[6] ...

The conversion involves usinglogical equivalences, such asdouble negation elimination,De Morgan's laws, and thedistributive law. Formulas built from theprimitiveconnectives${\displaystyle \{\wedge ,\vee ,\mathrm{\neg }\}}$^[7] can be converted to DNF by the followingcanonical term rewriting system:^[8]

The full DNF of a formula can be read off itstruth table.^[9][10] For example, consider the formula

${\displaystyle \varphi =((\mathrm{\neg }(p\wedge q))\leftrightarrow (\mathrm{\neg }r\uparrow (p\oplus q)))}$.^[11]

The correspondingtruth table is

${\displaystyle p}$	${\displaystyle q}$	${\displaystyle r}$		${\displaystyle (}$	${\displaystyle \mathrm{\neg }}$	${\displaystyle (p\wedge q)}$	${\displaystyle )}$	${\displaystyle \leftrightarrow }$	${\displaystyle (}$	${\displaystyle \mathrm{\neg }r}$	${\displaystyle \uparrow }$	${\displaystyle (p\oplus q)}$	${\displaystyle )}$
T	T	T			F	T		F		F	T	F
T	T	F			F	T		F		T	T	F
T	F	T			T	F		T		F	T	T
T	F	F			T	F		F		T	F	T
F	T	T			T	F		T		F	T	T
F	T	F			T	F		F		T	F	T
F	F	T			T	F		T		F	T	F
F	F	F			T	F		T		T	T	F

• The full DNF equivalent of${\displaystyle \varphi }$ is

${\displaystyle (p\wedge \mathrm{\neg }q\wedge r)\vee (\mathrm{\neg }p\wedge q\wedge r)\vee (\mathrm{\neg }p\wedge \mathrm{\neg }q\wedge r)\vee (\mathrm{\neg }p\wedge \mathrm{\neg }q\wedge \mathrm{\neg }r)}$

• The full DNF equivalent of${\displaystyle \mathrm{\neg }\varphi }$ is

${\displaystyle (p\wedge q\wedge r)\vee (p\wedge q\wedge \mathrm{\neg }r)\vee (p\wedge \mathrm{\neg }q\wedge \mathrm{\neg }r)\vee (\mathrm{\neg }p\wedge q\wedge \mathrm{\neg }r)}$

A propositional formula can be represented by one and only one full DNF.^[13] In contrast, severalplain DNFs may be possible. For example, by applying the rule${\displaystyle ((a\wedge b)\vee (\mathrm{\neg }a\wedge b))⇝b}$ three times, the full DNF of the above${\displaystyle \varphi }$ can be simplified to${\displaystyle (\mathrm{\neg }p\wedge \mathrm{\neg }q)\vee (\mathrm{\neg }p\wedge r)\vee (\mathrm{\neg }q\wedge r)}$. However, there are also equivalent DNF formulas that cannot be transformed one into another by this rule, see the pictures for an example.

It is a theorem that all consistent formulas inpropositional logic can be converted to disjunctive normal form.^{[14][15][16][17]} This is called theDisjunctive Normal Form Theorem.^{[14][15][16][17]} The formal statement is as follows:

Disjunctive Normal Form Theorem: Suppose${\displaystyle X}$ is a sentence in a propositional language${\displaystyle \mathcal{L}}$ with${\displaystyle n}$ sentence letters, which we shall denote by${\displaystyle A_1,...,A_n}$. If${\displaystyle X}$ is not a contradiction, then it is truth-functionally equivalent to a disjunction of conjunctions of the form${\displaystyle \pm A_1\wedge ...\wedge \pm A_n}$, where${\displaystyle +A_i=A_i}$, and${\displaystyle -A_i=\mathrm{\neg }{A}_i}$.^[15]

The proof follows from the procedure given above for generating DNFs fromtruth tables. Formally, the proof is as follows:

Suppose${\displaystyle X}$ is a sentence in a propositional language whose sentence letters are${\displaystyle A,B,C,\dots }$. For each row of${\displaystyle X}$'s truth table, write out a correspondingconjunction${\displaystyle \pm A\wedge \pm B\wedge \pm C\wedge \dots }$, where${\displaystyle \pm A}$ is defined to be${\displaystyle A}$ if${\displaystyle A}$ takes the value${\displaystyle T}$ at that row, and is${\displaystyle \mathrm{\neg }A}$ if${\displaystyle A}$ takes the value${\displaystyle F}$ at that row; similarly for${\displaystyle \pm B}$,${\displaystyle \pm C}$, etc. (thealphabetical ordering of${\displaystyle A,B,C,\dots }$ in the conjunctions is quite arbitrary; any other could be chosen instead). Now form thedisjunction of all these conjunctions which correspond to${\displaystyle T}$ rows of${\displaystyle X}$'s truth table. This disjunction is a sentence in${\displaystyle \mathcal{L}[A,B,C,\dots ;\wedge ,\vee ,\mathrm{\neg }]}$,^[18] which by the reasoning above is truth-functionally equivalent to${\displaystyle X}$. This construction obviously presupposes that${\displaystyle X}$ takes the value${\displaystyle T}$ on at least one row of its truth table; if${\displaystyle X}$ doesn’t, i.e., if${\displaystyle X}$ is acontradiction, then${\displaystyle X}$ is equivalent to${\displaystyle A\wedge \mathrm{\neg }A}$, which is, of course, also a sentence in${\displaystyle \mathcal{L}[A,B,C,\dots ;\wedge ,\vee ,\mathrm{\neg }]}$.^[15]

This theorem is a convenient way to derive many usefulmetalogical results in propositional logic, such as,trivially, the result that the set of connectives${\displaystyle \{\wedge ,\vee ,\mathrm{\neg }\}}$ isfunctionally complete.^[15]

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

There are${\displaystyle 2n}$ possible literals:${\displaystyle L=\{p_1,\mathrm{\neg }{p}_1,p_2,\mathrm{\neg }{p}_2,\dots ,p_n,\mathrm{\neg }{p}_n\}}$.

${\displaystyle L}$ has${\displaystyle (2^{2n}-1)}$ non-empty subsets.^[19]

This is the maximum number of conjunctions a DNF can have.^[13]

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:^[13]

${\displaystyle \begin{array}{l}(\mathrm{\neg }p)\vee (p)\vee (\mathrm{\neg }q)\vee (q)\vee \\ (\mathrm{\neg }p\wedge p)\vee \underset{\_}{(\mathrm{\neg }p\wedge \mathrm{\neg }q)}\vee \underset{\_}{(\mathrm{\neg }p\wedge q)}\vee \underset{\_}{(p\wedge \mathrm{\neg }q)}\vee \underset{\_}{(p\wedge q)}\vee (\mathrm{\neg }q\wedge q)\vee \\ (\mathrm{\neg }p\wedge p\wedge \mathrm{\neg }q)\vee (\mathrm{\neg }p\wedge p\wedge q)\vee (\mathrm{\neg }p\wedge \mathrm{\neg }q\wedge q)\vee (p\wedge \mathrm{\neg }q\wedge q)\vee \\ (\mathrm{\neg }p\wedge p\wedge \mathrm{\neg }q\wedge q)\end{array}}$

The longest possible full DNF has 4 conjunctions: they are underlined.

This formula is atautology. It can be simplified to${\displaystyle (\mathrm{\neg }p\vee p)}$ or to${\displaystyle (\mathrm{\neg }q\vee q)}$, which are also tautologies, as well as valid DNFs.

Example 2

Each DNF of the e.g. formula${\displaystyle (X_1\vee Y_1)\wedge (X_2\vee Y_2)\wedge \cdots \wedge (X_n\vee Y_n)}$ has${\displaystyle 2^n}$ conjunctions.

TheBoolean satisfiability problem onconjunctive normal form formulas isNP-complete. By theduality principle, so is the falsifiability problem on DNF formulas. Therefore, it isco-NP-hard to decide if a DNF formula is atautology.

Conversely, a DNF formula is satisfiable if, and only if, one of its conjunctions is satisfiable. This can be decided inpolynomial time simply by checking that at least one conjunction does not contain conflicting literals.

An important variation used in the study ofcomputational complexity isk-DNF. A formula is ink-DNF if it is in DNF and each conjunction contains at most k literals.^[20]

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

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• Sobolev, S.K. (2020) [1994],"Disjunctive normal form",Encyclopedia of Mathematics,EMS Press
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