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# (Distributive laws) For any three sets A, B, C prove that: I. A∪(B∩C)=(A∪B)∩(A∪C) [Distirbutive law of union over intersection] II. A∩[(B∪C)]=(A∩B)∪(A∩C) [Distributive law of intersection over union]

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## The correct option is I. Let x be an arbitary element of A∪(B∩C). Then, xϵA∪(B∩C)⇒ xϵA or xϵ(B∩C)⇒ xϵA or (xϵB) and xϵC⇒ (xϵA or xϵB) and (xϵA or xϵC) [∵ 'or' distributes 'and'] ⇒ xε(A∪B) and xϵ(A∪C).⇒ xϵ(A∪B)∩(A∪C) ∴ A∪(B∩C)⊆(A∪B)∩(A∪C) …(i) Again. let y be an arbitary element of (A∪B)∩(A∪C). Then, yϵ(A∪B)∩(A∪C)⇒yϵ(A∪B) and yϵ(A∪C)⇒(yϵA or yϵB) and (yϵA or yϵC)⇒yϵA or (yϵB and yϵC) [∵ 'or; distributes 'and'] ⇒ yϵA or yϵ(B∩C)⇒ yϵA∪(B∩C)∴ (A∪B)∩(A∪C)⊆A∪(B∩C) …(ii). From (i) and (ii), we get A∪(B∩C)=(A∪B)∩(A∪C) II. Let x be an arbitrary element of A∩(B∪C). Then, xϵA∩(B∪C)⇒xϵA and xϵ(B∪C)⇒xϵA and (xϵB or xϵC)⇒(xϵA and xϵB) or (xϵA and xϵC) [∵ 'and' distributes 'or'] ⇒ xϵ(A∩B) or xϵ(A∩C)⇒ xϵ(A∩B)∪(A∩C) ∴ A∩(B∪C)⊆(A∩B)∪(A∩C) …(iii) Again, let y be an arbitrary element of (A∩B)∪(A∩C). Then, yϵ(A∩B)∪(A∩C)⇒yϵ(A∩B) or yϵ(A∩C)⇒(yϵA and yϵB) or (yϵA and yϵC)⇒yϵA and (yϵB or yϵC) [∵ and' distributes 'or'] ⇒yϵA and yϵ(B∪C)⇒yϵA∩(B∪C). ∴ (A∩B)∪(A∩C)⊆A∩(B∪C) …(iv) From (iii) and (iv), we get A∩(B∪C)=(A∩B)∪(A∩C).  Suggest Corrections  15      Similar questions  Related Videos   Defining Hyperbola
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