Question

6 Show the following set identities, giren sets A, B, C, D.. ...a) ¢-C¢-A) = ADC 6). (A-BUA=A ...) An (B-6) = (ANB)-(ANG) d) (A-B). (B-A) = 0 e) (A-G) N (B-G) = CANB)- & .f) (B-A) (ANB) = 0 9) CAUB)-B = A-CARB) = A-B h) A-(B-C) = (A-B) U CANG) ..) (A-B)-C = A - (BUC). ..;) (A-B) CC-D) = (ANG)-(BUD).

Answer 1

(10) a) c- (C-A) = Ane at e-(c-A) y ace but xt(C-A) nec and a ca NE anc This e-((-A) = Anc . b) (A-B) UA - A ? x 4 (A – DU A + c (A - ) Y xe A <> DA. Las al So, . (A-BUA - A. FC) An (B-C) = (ANB) - Cane) ocet RECAMB)-Cane). <-> XE(ANB) ut && (AAC) <a> UEA and XEB but either a&A or ate <> REA and 2€(B-e) NE An (B-c) So, A0CB-C) = (ANB) – Cane) (A-B) OCB-A): ď . since since A, B A, B are a are given sets, either they have common elements or do not have common elements. for first case if ACB, then (A-BE & second care both (A-B) and (B-A) are emphy This for both the cases, we have (A-B) OCB-A) = $.

(C) (A -e) n (B-C) =(AMB) - C. Let ne (A-2) n (B-c) - REA and nEB but ade XE(ANB) but adc <=> x6 (ANB)-e. Thurs, we have, (A-C) (B-c)= (ANB)-c. (f) (B-ADO (ANB) - € Now, (B-A) n (ANR) = (Bra) CAMB) LA - complement of A] = BACARA OB = Bn&nB =¢ [ An A=&] (8) CAUB)-B = A- (ANB) = A-B (AUB)-B = (AUB) OBO - (ANBO) U (ORC) = (ANB) 04 PA-Ba Ange? - A-B Also A-(ANB) = An (Annje = ADCACUBE) - (ANA) O CANBC) • PU (ANB) = ANBC = A-B (h A-B-C) = (A-B) U (ANC) Let XE A-(B-c) (2 XEA but x& B-C <=> xEA, XEC but x&B XEA-B) and AXE (ANC)

=> 2 € (A-B) U (ANC) Therefore, we have, A- (0-2) = (A-BJU Cane) 0 (A-B)-C = A- (Bue) Let XE (A-B)-C <=> HELA-B) bat 2€ <> NEA but a&B and ade <> NEA but a¢ (BUL) =) € A-CBUC), Hence, we have (A-B)-C = A-CBUC). (j) (A-B)((-D) -(ANC) - (BUD). Let XE (A-B) n (C-D) AE (A-B) and at (C-1) XEA, KEC but a&B and a&D XE (ANC) but a&(BUD) La qe (ANC) - (BUD) Therefore, we get (A-Bn CC-D) = (Ane) - (BUD)
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