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MyWikiBiz, Author Your Legacy — Wednesday November 05, 2025
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The following equivalents can then be deduced:
The following equivalents may then be deduced:
{| 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==
