Question

De Morgan's laws for complementation: Let U be the universal set containing
finite sets A and B. Then (i) (A U B)' = A' B' (ii) (An B)' = A'U B'​

Answer 1

Step-by-step explanation:

Let P = (A U B)' and Q = A' ∩ B'

Let x be an arbitrary element of P then x ∈ P ⇒ x ∈ (A U B)'

⇒ x ∉ (A U B)

⇒ x ∉ A and x ∉ B

⇒ x ∈ A' and x ∈ B'

⇒ x ∈ A' ∩ B'

⇒ x ∈ Q

Therefore, P ⊂ Q …………….. (i)

Again, let y be an arbitrary element of Q then y ∈ Q ⇒ y ∈ A' ∩ B'

⇒ y ∈ A' and y ∈ B'

⇒ y ∉ A and y ∉ B

⇒ y ∉ (A U B)

⇒ y ∈ (A U B)'

⇒ y ∈ P

Therefore, Q ⊂ P …………….. (ii)

Now combine (i) and (ii) we get; P = Q i.e. (A U B)' = A' ∩ B'

Proof of De Morgan’s law: (A ∩ B)' = A' U B'

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

Therefore, 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

Therefore, N ⊂ M …………….. (ii)