Difference between revisions of "Exclusive disjunction"
MyWikiBiz, Author Your Legacy — Tuesday November 04, 2025
Jump to navigationJump to searchJon Awbrey (talk | contribs)  (add cats)  | 
				Jon Awbrey (talk | contribs)   (add cats)  | 
				||
| Line 66: | Line 66: | ||
[[Category:Computer Science]]  | [[Category:Computer Science]]  | ||
| + | [[Category:Formal Languages]]  | ||
| + | [[Category:Formal Sciences]]  | ||
| + | [[Category:Formal Systems]]  | ||
[[Category:Linguistics]]  | [[Category:Linguistics]]  | ||
[[Category:Logic]]  | [[Category:Logic]]  | ||
[[Category:Mathematics]]  | [[Category:Mathematics]]  | ||
| + | [[Category:Philosophy]]  | ||
[[Category:Semiotics]]  | [[Category:Semiotics]]  | ||
| − | |||
Revision as of 14:33, 21 May 2007
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 XOR q (also written as p + q, p ⊕ q, or p ≠ q) is as follows:
| p | q | p XOR q | 
|---|---|---|
| F | F | F | 
| F | T | T | 
| T | F | T | 
| T | T | F | 
The following equivalents can then be deduced:
\[\begin{matrix} p + q & = & (p \land \lnot q) & \lor & (\lnot p \land q) \\ \\ & = & (p \lor q) & \land & (\lnot p \lor \lnot q) \\ \\ & = & (p \lor q) & \land & \lnot (p \land q) \end{matrix}\]
See also
Logical operators
Related topics
Aficionados
- See Talk:Exclusive disjunction for discussions/comments regarding this article.
 - See Exclusive disjunction/Aficionados for those who have listed Exclusive disjunction as an interest.
 - See Talk:Exclusive disjunction/Aficionados for discussions regarding this interest.
 
<sharethis />