By R. Chuaqui

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So (*) R*(A U 8 ) C - (R*A) U (R*B) w i l l be shown. L e t us suppose t h a t ((R*A) u ( R * B ) ) n C = 0. Then, R * A n C = O = R * B n C . By (i), we o b t a i n A n R - l * C = O = BnR-'*C. Hence, ( A u 8) n R-'*C = 0. Using a g a i n (i), (R*(A U €4)) n C = 0. Now, i f we t a k e C = % ( ( R * A ) U ( R * B ) ) , (*) i s obtained. P R O O F OF ( v i ) . B)). Hence ( v i ) . P R O O F O F ( v i i ) . By ( i i i ) , R*(A%R-'*B) 5 R*A. Also, s i n c e R - l * B n = 0 we g e t from ( i ) , ( R * ( A % R - l * B ) ) n B = 0.

3. PROOF, (RoS) THEOREM, Assume R R2 -R C A S2 C S A R - 2 C _ R A S2 C -S A R o S oS = S OR = SoR. e. + (R oS)2 C - R oS. Then, 2 = R o ( S o R ) o S = ( R o R ) o ( S o S ) = R 2 O S2 R i s den6e, i f R C R -+ 5 R o S . R i s an 3 z(xRz A z R y ) ) . e. i f R = R - l and 2 R C R. There a r e s e v e r a l o t h e r ways o f s a y i n g t h a t R i s an equivalence relation: (i)says t h a t i f R i s symmetric and t r a n s i t i v e , then i t i s r e f l e x i v e . N o t i c e t h a t t h i s i s n o t t r u e f o r r e f l e x i v e i n A.

Hence, we a r e j u s t i f i e d i n u s i n g t h e same symbol f o r t h e union o r i n t e r s e c t i o n o f t h e elements o f a c l a s s and t h e corresponding o p e r a t o r f o r t h e elements o f a c o l l e c t i o n . 5 THEOREM I ( i ) A u B = U { X : X = A V X = B) (ii) Ant3 = n { X : X = A V X = B) . Therefore, theorems about g e n e r a l i z e d unions o r i n t e r s e c t i o n i n c l u d e as p a r t i c u l a r cases t h e corresponding ones about t h e b i n a r y o p e r a t i o n .