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Exclusive disjunction
(view source)
Revision as of 16:36, 13 May 2012
55 bytes added
,
16:36, 13 May 2012
spacing
Line 25:
Line 25:
<br>
<br>
−
The following equivalents
can
then be deduced:
+
The following equivalents
may
then be deduced:
−
:
<math>\begin{matrix}
+
{| align="center" cellspacing="10" width="90%"
−
p + q & = & (p \land \lnot q) & \lor & (\lnot p \land q)
\\
+
|
−
\\
+
<math>\begin{matrix}
−
& = & (p \lor q) & \land & (\lnot p \lor \lnot q)
\\
+
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)
& = & (p \lor q) & \land & \lnot (p \land q)
\end{matrix}</math>
\end{matrix}</math>
+
|}
==Syllabus==
==Syllabus==
Jon Awbrey
12,080
edits