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"¯Œ^ ƒc[ƒuƒƒbƒN —Žq-Dec 21,  · 51 Sets and Operations on Sets Before beginning this section, it would be a good idea to review sets and set notation, including the roster method and set builder notation, in Section 23 In Section 21, we used logical operators (conjunction, disjunction, negation) to form new statements from existing statementsAbout Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How works Test new features Press Copyright Contact us Creators

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First find set A U B This is just the set of A and B combined and any duplicates are tossed A U B = {q, s, u, w, y} U {q, s, y, z} = {q, s, u, w, y, z}N(AuBuC) = n(A)n(B) nn(AnB)n(Anc)n(BnC)n(AnBnC) As we know n(AuB)=n(A) n(B)n(AnB) Here when we add A and B we are adding (AnB) two times , as it is present in both A and B So we subtract (AnB) one time from A and B Which means we needQuestion Show (AB) U (BC) = (A U B) (B N C)Proof (⊆, Element Argument)Let X € (AB) U (BC) € = ElementThen X € A And X ~€ B, Or X € B And X ~€ C Consider Thefirst Case, So Suppose X € A And X ~€ B Since X € A, X € (A U B)Since X ~€ B, And (B N C) ⊆ B, The X ~€ (B N C)Thus X € (A U B) (B N C) Consider The Casethat X

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