By Leopoldo Nachbin (Eds.)

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**Example text**

T h i s i s e q u i v a l e n t t o f? f S A R' f S ' . Taking A €3 = ( A €3) U ( 8 A ) , R f S, R' # S' i s c l e a r l y e q u i v a l e n t t o R I S f 0 A R ' L S' # 0 . 5 ( x i i i ) , t h i s i s transformed i n t o ( V x V ) o ( R I S ) o ( V x V ) = V x V A ( V x V ) o ( R ' L S ' ) o ( V x V ) = V x V; and t h i s c o n j u n c t i o n i s e q u i v a l e n t t o - ( 2 ) ( ( V x V ) o ( R IS) - o (V x V ) ) n ( ( V x V ) O ( R ' I S ' ) o ( V x V)) = 36 ROLAND0 CHUAQUI = vx V . e. t o (3) - ( ( ( V x V ) o (RIS) 0 (Vx V))) Let T - = o ( V X V ) ) n ( ( V XV ) O (R’ 2 S ’ ) o 0.

Using Ax Ref we o b t a i n , 3u( Uu c -u A 3x 3y (x,y 5u A x # y A x,y E A n u ) ) . Hence, t h e r e i s a t r a n s i t i v e s e t u, such t h a t , (1) 3 x 3 y ( x # y A x,y E A n u). Therefore, s i n c e A n u -A C = ( a , b } , A n u = 0, A n u = { a } , An u = (6) 32 ROLAND0 C H U A Q U I o r Anu = { a , b } . I n t h e f i r s t t h r e e cases, (1) would be false. ( 2 ) Anu = { a , b l C u . Applying, now, Ax Sub we o b t a i n t h a t {a,bl tive E V. e. i s obtained. THEOREM ( B I- Ax Num).

E. U A C -A ) . V x V , 0, I D , and D v a r e symmetric, w h i l e t h e o t h e r s a r e not. 2 PROOF, theorem. q r z. THEOREM SCHEMA, Let Suppose Hence x r z . R n '0 0 . xRq A yRz. Therefore, le,t r be a tehm and 4 a jjotvnuh. Then, { r : @ I and assume t h e hypotheses o f t h e Xn-1 Then f o r a l l Xo,... Xnml, such t h a t 4 , x r q A x R z. 43 AXIOMATIC SET THEORY The c o m p o s i t i o n o f two t r a n s i t i v e r e l a t i o n s i s n o t , i n genera1,trans i t i v e . 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.