Question

Prove that A ∩ (A∪B)’ = ϕ

Answer 1

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