MyWikiBiz, Author Your Legacy — Wednesday April 02, 2025
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55 bytes added
, 16:36, 13 May 2012
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<br>
<br>
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The following equivalents can then be deduced:
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The following equivalents may then be deduced:
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: <math>\begin{matrix}
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{| align="center" cellspacing="10" width="90%"
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p + q & = & (p \land \lnot q) & \lor & (\lnot p \land q) \\
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|
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\\
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<math>\begin{matrix}
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& = & (p \lor q) & \land & (\lnot p \lor \lnot q) \\
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p + q & = & (p \land \lnot q) & \lor & (\lnot p \land q)
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\\
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\\[6pt]
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& = & (p \lor q) & \land & (\lnot p \lor \lnot q)
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\\[6pt]
& = & (p \lor q) & \land & \lnot (p \land q)
& = & (p \lor q) & \land & \lnot (p \land q)
\end{matrix}</math>
\end{matrix}</math>
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|}
==Syllabus==
==Syllabus==