For any two sets A and B, prove that
(i) (A ∪ B) – B = A – B
(ii) A – (A ∩ B) = A – B
(iii) A – (A – B) = A ∩ B
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(i) (A ∪ B) – B = A – B
Consider LHS (A ∪ B) – B
= (A – B) ∪ (B – B)
= (A – B) ∪ ϕ (here, B – B = ϕ)
= A – B (here, x ∪ ϕ = x for any set)
= RHS
Thus proved.
(ii) A – (A ∩ B) = A – B
Consider LHS A – (A ∩ B)
= (A – A) ∩ (A – B)
= ϕ ∩ (A – B) (here, A - A = ϕ)
= A – B
= RHS
Thus proved.
(iii) A – (A – B) = A ∩ B
Consider LHS A – (A – B)
Suppose, x ∈ A – (A – B) = x ∈ A and x ∉ (A – B)
x ∈ A and x ∉ (A ∩ B)
= x ∈ A ∩ (A ∩ B)
= x ∈ (A ∩ B)
= (A ∩ B)
= RHS
Thus proved.
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