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de Morgan's Laws

Let $\cup$ represent ``or'', $\cap$ represent ``and'', and $'$ represent ``not.'' Then, for two logical units $E$ and $F$,

\begin{displaymath}
(E\cup F)' = E'\cap F'
\end{displaymath}


\begin{displaymath}
(E\cap F)' = E'\cup F'.
\end{displaymath}




© 1996-9 Eric W. Weisstein
1999-05-24