12,080
edits
Changes
MyWikiBiz, Author Your Legacy — Friday November 08, 2024
Jump to navigationJump to searchspacing
<br>
<br>
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==
