1.

Prove De Morgan’s law for (A∩ B)ʹ = Aʹ ∪ Bʹ​

Answer»

Hey friend!

Here is your ANSWER:

To prove De Morgan's Law for ( A ∩ B )' = A' ∪ B'

Proof:

Let M = (A ∩ B)' and N = A' U B'

Let x be an arbitrary element of M

then x ∈ M ⇒ x ∈ (A ∩ B)'

⇒ x ∉ (A ∩ B)

⇒ x ∉ A or x ∉ B

⇒ x ∈ A' or x ∈ B'

⇒ x ∈ A' U B'

⇒ x ∈ N

⇔ M ⊂ N …………….. (i)

Again,

Let y be an arbitrary element of N

then y ∈ N ⇒ y ∈ A' U B'

⇒ y ∈ A' or y ∈ B'

⇒ y ∉ A or y ∉ B

⇒ y ∉ (A ∩ B)

⇒ y ∈ (A ∩ B)'

⇒ y ∈ M

⇔ N ⊂ M …………….. (II)

Now,

From (i) and (ii) we GET;

→  M = N

→ (A ∩ B)' = A' U B'

Hence proved

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