LHS = A ∩ (A ⋃ B)’

Using De-Morgan’s law (A ⋃ B)’ = (A’ ∩ B’)

⇒ LHS = A ∩ (A’ ∩ B’)

⇒ LHS = (A ∩ A’) ∩ (A ∩ B’)

We know that A ∩ A’ = ϕ

⇒ LHS = ϕ ∩ (A ∩ B’)

We know that intersection of null set with any set is null set only

Let (A ∩ B’) be any set X hence

⇒ LHS = ϕ ∩ X

⇒ LHS = ϕ

⇒ LHS = RHS

Hence proved

Answered by Sakshi | 11 months agoFind the symmetric difference A Δ B, when A = {1, 2, 3} and B = {3, 4, 5}.