De Morgan's Law

• August De Morgan $(1806-1871)$ was a British mathematician who formulated De Morgan Law
• If $A$ and $B$ are the subsets of a universal set $U$, then

$i)$  $(A\cup B)'$ $=(A'\cap B')$            $ii)$  $(A\cap B)'$ $=(A'\cup B')$

$i)$  $(A\cup B)'$ $=(A'\cap B')$

Proof:
Let  $x\epsilon (A\cup B)'$

$x\notin (A\cup B)$

$x\notin A$  and  $x\notin B$

$x\epsilon (A'\cap B')$

$(A\cup B)'\subseteq (A'\cap B')$      $(i)$

Let  $y\epsilon (A\cup B)'$

$y\notin (A\cup B)$

$y\notin A$   and   $y\notin B$

$y\epsilon (A'\cap B')$

$(A\cup B)'\subseteq (A'\cap B')$      $(ii)$

From $(i)$, and $(ii)$ we have,

$(A\cup B)'$ $=A'\cap B'$

Similarly, we can prove. $(A\cap B)'$ $=A'\cup B'$