Solution :
We have to prove that [tex]$\overline{A \cup B} = \overline{A} \cap \overline{B}$[/tex] (De-Morgan's law)
Let [tex]$x \in \bar{A} \cap \bar{B}, $[/tex] then [tex]$x \in \bar{A}$[/tex] and [tex]$x \in \bar{B} $[/tex]
and so [tex]$x \notin \bar{A}$[/tex] and [tex]$x \notin \bar{B} $[/tex].
Thus, [tex]$x \notin A \cup B$[/tex] and so [tex]$x \in \overline{A \cup B}$[/tex]
Hence, [tex]$\bar{A} \cap \bar{B} \subset \overline{A \cup B}$[/tex] .........(1)
Now we will show that [tex]$\overline{A \cup B} \subset \overline{A} \cap \overline{B}$[/tex]
Let [tex]$x \in \overline{A \cup B}$[/tex] ⇒ [tex]$x \notin A \cup B$[/tex]
Thus x is present neither in the set A nor in the set B, so by definition of the union of the sets, by definition of the complement.
[tex]$x \in \overline{A}$[/tex] and [tex]$x \in \overline{B}$[/tex]
Therefore, [tex]$x \in \overline{A} \cap \overline{B}$[/tex] and we have [tex]$\overline{A \cup B} \subset \overline{A} \cap \overline{B}$[/tex] .............(2)
From (1) and (2),
[tex]$\overline{A \cup B} = \overline{A} \cap \overline{B}$[/tex]
Hence proved.
