# Prove both the DeMorgan’s laws using truth tables?

Mar 12, 2018

See below

#### Explanation:

Let $x \in {\left(A \cup B\right)}^{c}$ then $x \notin A \cup B$. That means $x \notin A \mathmr{and} x \notin B$. This is the same to say $x \in {A}^{c} \mathmr{and} x \in {B}^{c}$. We can write both conditions in that way: $x \in {A}^{c} \cap {B}^{c}$

Thus, the inclusion ${\left(A \cup B\right)}^{c} \subset {A}^{c} \cap {B}^{c}$ (1) is true.

Lets see the oposite. That will prove ${\left(A \cup B\right)}^{c} = {A}^{c} \cap {B}^{c}$

Let $x \in {A}^{c} \cap {B}^{c}$. That means $x \in {A}^{c} \mathmr{and} x \in {B}^{c}$. This it's the same to say $x \notin A \mathmr{and} x \notin B$. Then $x \notin A \cup B$. And finally, that it's the same to say $x \in {\left(A \cup B\right)}^{c}$

We have prove ${\left(A \cup B\right)}^{c} \supset {A}^{c} \cap {B}^{c}$ (2)

Because (1) and (2) are true simultaneusly, then ${\left(A \cup B\right)}^{c} = {A}^{c} \cap {B}^{c}$ QED