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Disjunction[expr,{a1,a2,…}]

gives the disjunction of expr over all choices of the Boolean variables ai.

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Disjunction

Disjunction[expr,{a1,a2,…}]

gives the disjunction of expr over all choices of the Boolean variables ai.

Details

Examples

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Basic Examples  (3)

The disjunction over a set of variables:

Wolfram Language code: Disjunction[f[a, b], {a, b}]

Check whether an expression is satisfiable:

Wolfram Language code: Disjunction[Xor[a, b, c], {a, b, c}]

Find the conditions on a for a⊽b to be satisfiable:

Wolfram Language code: Disjunction[Nor[a, b], {b}]

Properties & Relations  (5)

Disjunction effectively computes the Or over all truth values of the listed variables:

Wolfram Language code: Or@@BooleanTable[a && b && (c || d), {c, d}]
Wolfram Language code: Disjunction[a && b && (c || d), {c, d}]
Wolfram Language code: TautologyQ[Equivalent[%%, %]]

Disjunction is typically more efficient and can work large numbers of variables:

Wolfram Language code: f = BooleanConvert[BooleanCountingFunction[{30, 70}, 100], "BFF"]
Wolfram Language code: Disjunction[f@@Array[x, 100], Array[x, 20]]

Disjunction eliminates ∃ (Exists) quantifiers for the list of variables:

Wolfram Language code: Resolve[Exists[{c, d}, a && b && (c || d)]]
Wolfram Language code: Disjunction[a && b && (c || d), {c, d}]

Use Resolve to eliminate more general combinations of quantifiers:

Wolfram Language code: Resolve[Subscript[∀, a]Subscript[∃, b]Xor[a, b, c]]

SatisfiableQ is Disjunction over all variables:

Wolfram Language code: SatisfiableQ[¬Implies[Implies[a, b]∧a, b]]
Wolfram Language code: Disjunction[¬Implies[Implies[a, b]∧a, b], {a, b}]

Use Conjunction to compute And over a list of variables:

Wolfram Language code: Conjunction[f[a, b, c], {b, c}]

Conjunction is related to Disjunction by de Morgan's law:

Wolfram Language code: TautologyQ[Equivalent[¬Disjunction[¬f[a, b], {a, b}], Conjunction[f[a, b], {a, b}]]]

Disjunction is effectively repeated Or, just as Sum is repeated Plus:

Wolfram Language code: Sum[f[i, j], {i, 0, 1}, {j, 0, 1}]
Wolfram Language code: Disjunction[f[i, j], {i, j}]

Represent Disjunction in terms of Sum:

Wolfram Language code: Or@@Sum[f[i, j], {i, {True, False}}, {j, {True, False}}]
Wolfram Research (2008), Disjunction, Wolfram Language function, https://reference.wolfram.com/language/ref/Disjunction.html.

Text

Wolfram Research (2008), Disjunction, Wolfram Language function, https://reference.wolfram.com/language/ref/Disjunction.html.

CMS

Wolfram Language. 2008. "Disjunction." Wolfram Language & System Documentation Center. Wolfram Research. https://reference.wolfram.com/language/ref/Disjunction.html.

APA

Wolfram Language. (2008). Disjunction. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/Disjunction.html

BibTeX

@misc{reference.wolfram_2026_disjunction, author="Wolfram Research", title="{Disjunction}", year="2008", howpublished="\url{https://reference.wolfram.com/language/ref/Disjunction.html}", note=[Accessed: 13-June-2026]}

BibLaTeX

@online{reference.wolfram_2026_disjunction, organization={Wolfram Research}, title={Disjunction}, year={2008}, url={https://reference.wolfram.com/language/ref/Disjunction.html}, note=[Accessed: 13-June-2026]}