De Morgan's second law

• October 7th 2008, 01:00 PM
lds09
De Morgan's second law
Hi,

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

Thanks!
• October 7th 2008, 01:27 PM
Plato
I do not know the why you have them numbered.
Here is a possiblity.
$\begin{gathered}
\left( {A^c \cup B^c } \right)^c = \left( {A^c } \right)^c \left( {B^c } \right)^c = AB \hfill \\
\left( {AB} \right)^c = A^c \cup B^c \hfill \\
\end{gathered}$
• October 7th 2008, 01:29 PM
Moo
Hello,
Quote:

Originally Posted by lds09
Hi,

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

Thanks!

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

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

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

Take the complementary of both sides :

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

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

This is the second law.