# Thread: De Morgan's second law

1. ## De 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!

2. I do not know the why you have them numbered.
Here is a possiblity.
$\displaystyle \begin{gathered} \left( {A^c \cup B^c } \right)^c = \left( {A^c } \right)^c \left( {B^c } \right)^c = AB \hfill \\ \left( {AB} \right)^c = A^c \cup B^c \hfill \\ \end{gathered}$

3. Hello,
Originally Posted by lds09
Hi,

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

Thanks!
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.