Question -
Answer -
(i) A ∩ (A’ ∪ B) = A ∩ B
Let us consider LHS A ∩ (A’ ∪ B)
Expanding
(A ∩ A’) ∪ (A ∩ B)
We know, (A ∩ A’) =ϕ
⇒ ϕ ∪ (A∩ B)
⇒ (A ∩ B)
∴ LHS = RHS
Hence proved.
(ii) A – (A – B) = A ∩ B
For any sets A and B we have De-Morgan’s law
(A ∪ B)’ = A’ ∩ B’, (A ∩ B) ‘ = A’ ∪ B’
Consider LHS
= A – (A–B)
= A ∩ (A–B)’
= A ∩ (A∩B’)’
= A ∩ (A’ ∪ B’)’) (since, (B’)’ = B)
= A ∩ (A’ ∪ B)
= (A ∩ A’) ∪ (A ∩ B)
= ϕ ∪ (A ∩ B) (since, A ∩ A’ = ϕ)
= (A ∩ B) (since, ϕ ∪ x = x, for any set)
= RHS
∴ LHS=RHS
Hence proved.
(iii) A ∩ (A ∪ B’) = ϕ
Let us consider LHS A ∩ (A ∪ B’)
= A ∩ (A ∪ B’)
= A ∩ (A’∩ B’) (By De–Morgan’s law)
= (A ∩ A’) ∩ B’ (since, A ∩ A’ = ϕ)
= ϕ ∩ B’
= ϕ (since, ϕ ∩ B’ = ϕ)
= RHS
∴ LHS=RHS
Hence proved.
(iv) A – B = A Δ (A ∩ B)
Let us consider RHS A Δ (A ∩ B)
A Δ (A ∩ B) (since, E Δ F = (E–F) ∪ (F–E))
= (A – (A ∩ B)) ∪ (A ∩ B –A) (since, E – F = E ∩ F’)
= (A ∩ (A ∩ B)’) ∪ (A ∩ B ∩ A’)
= (A ∩ (A’ ∪ B’)) ∪ (A ∩ A’ ∩ B) (by using De-Morgan’s law and associative law)
= (A ∩ A’) ∪ (A ∩ B’) ∪ (ϕ ∩ B) (by using distributive law)
= ϕ ∪ (A ∩ B’) ∪ ϕ
= A ∩ B’ (since, A ∩ B’ = A–B)
= A – B
= LHS
∴ LHS=RHS
Hence Proved