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DaniilM [7]
3 years ago
9

Prove De Morgan's law by showing that each side is a subset of the other side by considering x ∈ A⎯⎯⎯ A ¯ ∩ B⎯⎯⎯ B ¯ .

Mathematics
1 answer:
adelina 88 [10]3 years ago
3 0

Solution :

We have to prove that $\overline{A \cup B} = \overline{A} \cap \overline{B}$   (De-Morgan's law)

Let  $x \in \bar{A} \cap \bar{B}, $ then $x \in \bar{A}$ and $x \in \bar{B} $

and so $x \notin \bar{A}$ and $x \notin \bar{B} $.

Thus, $x \notin A \cup B$ and so $x \in \overline{A \cup B}$

Hence, $\bar{A} \cap \bar{B} \subset \overline{A \cup B}$   .........(1)

Now we will show that $\overline{A \cup B} \subset \overline{A} \cap \overline{B}$

Let $x \in \overline{A \cup B}$ ⇒ $x \notin A \cup B$

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.

$x \in \overline{A}$ and  $x \in \overline{B}$

Therefore, $x \in \overline{A} \cap \overline{B}$ and we have $\overline{A \cup B} \subset \overline{A} \cap \overline{B}$  .............(2)

From (1) and (2),

$\overline{A \cup B} = \overline{A} \cap \overline{B}$

Hence proved.

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