DeMorgan’s Second Law

DeMorgan’s second law is about the complement of the intersection of sets.

Complement of the intersection of sets

If we want to take the complement of a set defined in terms of set intersections, DeMorgan’s second laws a way to do that. For any arbitrary sets $A$ and $B$, if we want to compute $\overline{(A\cap B)}$, it isn’t equal to $\overline{A} \cap \overline{B}$. We can demonstrate this fact with the help of an example.

Let us take some sets to make a concrete example. We assume that $\mathbb{U}=\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $A=\{2, 3, 4, 5, 6\}$, and $B=\{1, 3, 5, 7\}$. From this information, we can derive the following:

$A\cap B = \{3, 5\}$

$\overline{A} =\{0, 1, 7, 8, 9\}$

$\overline{B}=\{0, 2, 4, 6, 8, 9\}$

$\overline{A \cap B} = \{0, 1, 2, 4, 6, 7, 8, 9\}$

$\overline{A}\cap \overline{B} = \{0, 8, 9\}$

As a result, it’s evident that $\overline{(A\cap B)} \ne \overline{A} \cap \overline{B}$.

For any arbitrary sets $A$ and $B$, DeMorgan’s second law tells us that $\overline{(A\cap B)}= \overline{A} \cup \overline{B}$.

Let’s see the argument for why this is true. First, we show that $\overline{(A\cap B)}\sube (\overline{A} \cup \overline{B})$.