Hi,

how do I prove De Morgan's second law $\displaystyle (AB)^c = A^c \cup B^c $ by using De Morgan's first law: $\displaystyle (E \cup F)^c = E^c F^c $

Thanks!

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- Oct 7th 2008, 12:00 PMlds09De Morgan's second law
Hi,

how do I prove De Morgan's second law $\displaystyle (AB)^c = A^c \cup B^c $ by using De Morgan's first law: $\displaystyle (E \cup F)^c = E^c F^c $

Thanks! - Oct 7th 2008, 12:27 PMPlato
I do not know the why you have them numbered.

Here is a possiblity.

$\displaystyle \begin{gathered}

\left( {A^c \cup B^c } \right)^c = \left( {A^c } \right)^c \left( {B^c } \right)^c = AB \hfill \\

\left( {AB} \right)^c = A^c \cup B^c \hfill \\

\end{gathered} $ - Oct 7th 2008, 12:29 PMMoo
Hello,

Start from de Morgan's first law for $\displaystyle A^c$ and $\displaystyle B^c$ :

$\displaystyle (A^c \cup B^c)^c=(A^c)^c \cap (B^c)^c$

$\displaystyle (A^c \cup B^c)^c=A \cap B$ (by noting that $\displaystyle (A^c)^c=A$

Take the complementary of both sides :

$\displaystyle \underbrace{\bigg((A^c \cup B^c)^c\bigg)^c}_{=A^c \cup B^c}=\bigg(A \cap B\bigg)^c$

$\displaystyle A^c \cup B^c=(A \cap B)^c$

This is the second law.