Difference between revisions of "Exclusive disjunction"

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* [http://mywikibiz.com/Exclusive_disjunction Exclusive Disjunction @ MyWikiBiz]
 
* [http://mywikibiz.com/Exclusive_disjunction Exclusive Disjunction @ MyWikiBiz]
* [http://mathweb.org/wiki/Exclusive_disjunction Exclusive Disjunction @ MathWeb Wiki]
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* [http://intersci.ss.uci.edu/wiki/index.php/Exclusive_disjunction Exclusive Disjunction @ InterSci Wiki]
* [http://netknowledge.org/wiki/Exclusive_disjunction Exclusive Disjunction @ NetKnowledge]
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* [http://wiki.oercommons.org/mediawiki/index.php/Exclusive_disjunction Exclusive Disjunction @ OER Commons]
 
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* [http://wiki.oercommons.org/mediawiki/index.php/Exclusive_disjunction Exclusive Disjunction @ OER Commons]
 
 
* [http://p2pfoundation.net/Exclusive_Disjunction Exclusive Disjunction @ P2P Foundation]
 
* [http://p2pfoundation.net/Exclusive_Disjunction Exclusive Disjunction @ P2P Foundation]
 
* [http://semanticweb.org/wiki/Exclusive_disjunction Exclusive Disjunction @ SemanticWeb]
 
* [http://semanticweb.org/wiki/Exclusive_disjunction Exclusive Disjunction @ SemanticWeb]
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* [http://beta.wikiversity.org/wiki/Exclusive_disjunction Exclusive Disjunction @ Wikiversity Beta]
 
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Revision as of 03:54, 14 May 2012

This page belongs to resource collections on Logic and Inquiry.

Exclusive disjunction, also known as logical inequality or symmetric difference, is an operation on two logical values, typically the values of two propositions, that produces a value of true just in case exactly one of its operands is true.

The truth table of \(p ~\operatorname{XOR}~ q\) (also written as \(p + q\!\) or \(p \ne q\!\)) is as follows:


\(\text{Exclusive Disjunction}\!\)
\(p\!\) \(q\!\) \(p ~\operatorname{XOR}~ q\)
\(\operatorname{F}\) \(\operatorname{F}\) \(\operatorname{F}\)
\(\operatorname{F}\) \(\operatorname{T}\) \(\operatorname{T}\)
\(\operatorname{T}\) \(\operatorname{F}\) \(\operatorname{T}\)
\(\operatorname{T}\) \(\operatorname{T}\) \(\operatorname{F}\)


The following equivalents may then be deduced:

\(\begin{matrix} p + q & = & (p \land \lnot q) & \lor & (\lnot p \land q) \\[6pt] & = & (p \lor q) & \land & (\lnot p \lor \lnot q) \\[6pt] & = & (p \lor q) & \land & \lnot (p \land q) \end{matrix}\)

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Portions of the above article were adapted from the following sources under the GNU Free Documentation License, under other applicable licenses, or by permission of the copyright holders.

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