# Timeless Theorems of Mathematics/De Morgan's laws

**De Morgan\'s Law** is a fundamental principle in Logic and Set Theory.
It establishes a useful relationship between the logical operators
\'AND\' and \'OR\' when negations *(NOT)* are applied. There are two
primary forms of De Morgan\'s Law, known as De Morgan\'s First Law and
De Morgan\'s Second Law. These laws state that,

1.  The negation of a disjunction is the conjunction of the negations,
    $(A \cup B)' = A' \cap B'$
2.  The negation of a conjunction is the disjunction of the negations,
    $(A \cap B)' = A' \cup B'$

## Proof

!De Morgan\'s Law for sets $A$ and
$B$ Let $A$ and
$B$ are two sets. De Morgan\'s First Law states that the complement of
the union of $A$ and $B$ sets is the same as the intersection of the
complements of $A$ and $B$. That means, $(A \cup B)' = A' \cap B'$, or
$\overline {(A\cdot B)} = (\overline {A} + \overline {B})$. And De
Morgan\'s First Law states that the complement of the intersection of
$A$ and $B$ is the same as the union of their complements,
$(A \cap B)' = A' \cup B'$, or
$\overline {(A+B)} = (\overline {A} \cdot \overline {B})$.

### De Morgan\'s First Law

Assume $x \in (A\cup B)'$. Then $x \not \in A \cup B$.

$\implies x \not \in A$ AND $x \not \in B$

$\implies x \in A'$ AND $x \in B'$

$\implies x \in (A'\cap B')$

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

Again, Let, $x \in (A'\cap B')$. Then, $x \in A'$ AND $x \in B'$

$\implies x \not \in A$ AND $x \not \in B$

$\implies x \not \in A \cup B$.

$\implies x \in (A\cup B)'$

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

Therefore, $(A \cap B)' = A' \cup B'$ ***\[Proved\]***

### De Morgan\'s Second Law

Assume $x \in (A\cap B)'$. Then $x \not \in A \cap B$.

$\implies x \not \in A$ OR $x \not \in B$

$\implies x \in A'$ OR $x \in B'$

$\implies x \in (A'\cup B')$

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

Again, Let, $x \in (A'\cup B')$. Then, $x \in A'$ OR $x \in B'$

$\implies x \not \in A$ OR $x \not \in B$

$\implies x \not \in A \cap B$.

$\implies x \in (A\cap B)'$

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

Therefore, $(A \cap B)' = A' \cup B'$ ***\[Proved\]***

