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
(iv) A∪(B−A)=A∪B
(v) (A−B)∪(A∩B)=A
(i) (A∪B)−B=A−B
LHS = (A∪B)−B=(A∪B)∩B′ [ ∵A−B=A∩B′]
(A∩B′)∪(B∩B′)=(A∩B′)∪ϕ [∵B∩B′=ϕ]
= A∩B′
= A- B = RHS
(ii) \( A - (A \cap B) = A - B
LHS=A−(A∩B)=A∩(A∩B)′ [∵A−B=A∩B′]
= A∩(A′∪B′) [∵(A∩B)′=A′∪B′]
= (A∩A′)∪(A∩B′)=ϕ∪(A∩B′)
= A∩B′ [∵ϕ∪A=A]
= A - B = RHS
(iii)A−(A−B)=A∩B
LHS =A−(A−B)=A−(A∩B′)
[∵A−B=A∩B′]
= A∩(A∩B′)′=A∩(A′∪(B′)′] [∵(A∩B)′=A′∪B′]
= A∩(A′∪B) [∵(A′)′=A]
= (A∩A′)∪(A∩B)=ϕ∪(A∩B)
= A∩B = RHS
(iv) A∪(B−A)=A∪B
LHS = A∪(B−A)=A∪(B∩A′) [∵A−B=A∩B′]
= (A∪B)∩(A∪A′)=(A∪B)∩U [∵A∪A′=U]
= A∪B=RHS [∵A∩U=A]
(v) (A−B)∪(A∩B)=A
LHS =(A−B)∪(A∩B)
= [(A−B)∪[(A−B)∪B]
= A∩(A∪B)=A=RHS