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!
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} $
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.