Solve this : Prove using properties of sets and their complements . ( 1 ) A ∪ B ∩ A ∪ B ' = A ( 2 ) A - A ∩ B = A - B ( 3 ) A ∪ B - C = A - C ∪ B - C ( 4 ) A - B ∪ C = A - B ∩ A - C ( 5 ) A ∩ B - C = A ∩ B - A ∩ C . Share with your friends Share 3 Lovina Kansal answered this Dear student (3) A∪B-C=A∪B∩C'=(A∩C')∪(B∩C')=(A-C)∪(B-C) (4) A-B∪C=A-B∩A-CLHS=A-B∪C=A∩B∪C' ∵X-Y=X∩Y'=A∩(B'∩C') [∵(B∪C)'=B'∩C']=(A∩B')∩(A∩C')=(A-B)∩(A-C)=RHS(5) A∩(B-C)=(A∩B)-A∩CLet x be any arbitrary element of A∩(B-C).Thenx∈A∩(B-C)⇒x∈A and x∈B-C⇒x∈A and x∈B and x∉C⇒x∈A and x∈B and x∈A and x∉C⇒x∈(A∩B) and x∉A∩C⇒x∈(A∩B)-A∩CSo, A∩(B-C)⊆(A∩B)-A∩CSimilarly, (A∩B)-A∩C⊆A∩(B-C)Hence,A∩(B-C)=(A∩B)-A∩CSimilarly try remaining parts. Regards 0 View Full Answer