Disjunct Normal Form

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.

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. 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 ($\wedge$), or ($\vee$), and not ($\neg$). 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:
# ''DNF'' → (''Conjunction'') $\vee$ ''DNF''
# ''DNF'' → (''Conjunction'')
# ''Conjunction'' → ''Literal'' $\wedge$ ''Conjunction''
# ''Conjunction'' → ''Literal''
# ''Literal'' → $\neg$''Variable''
# ''Literal'' → ''Variable''
Where ''Variable'' is any variable.

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

*$(A \land \neg B \land \neg C) \lor (\neg D \land E \land F)$
*$(A \land B) \lor (C)$
*$(A \land B)$
*$(A)$

However, the following formulas are not in DNF:
*$\neg(A \lor B)$, since an OR is nested within a NOT
*$\neg(A \land B) \lor C$, since an AND is nested within a NOT
*$A \lor (B \land (C \lor D))$, since an OR is nested within an AND

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

Conversion to DNF

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 $(X_1 \lor Y_1) \land (X_2 \lor Y_2) \land \dots \land (X_n \lor Y_n)$ to DNF yields a formula with 2^''n'' terms.

Every particular Boolean function can be represented by one and only oneIgnoring variations based on associativity and commutativity of AND and OR. ''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.

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.

See also

* 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

Notes

References

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External links

* {{springer, title=Disjunctive normal form, id=p/d033300

Normal forms (logic)