By R. Chuaqui

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R i s not enough t o determine A ; because A might be d i f f e r e n t from D R. W i t h j u s t R we could not have superclasse A with A ( O ) , because f o r a l l x E D R, R*{x} f 0. 3 THEOREM SCHEMA, L e t F be a u m q aperraLLiun. Then, Wx(xEB + F ( x ) = 0 ) + [ F (x) : x E A ] = [F(x) : x e A U B ] AXIOMATIC S E T T H E O R Y PROOF, Suppose t h a t F ( x ) = 0, f o r X E 8. 41 We have, [F(x) : x g A u B ] = u (F(x) x ( x } :x E A u B ) = B u t , by hypothesis, Therefore, : x E B ) = 0. u { F ( x ) x { x ] :x E A } U u { F ( x ) x { X I :x E B I F(x) { x } = 0, for x € B .

Ii) RIB = R B . (iii) R*A = n v x Cy . : 3 x ( x € A A x R y ) } AIR i s R r e s t r i c t e d i n i t s range t o A ; R I B i s R r e s t r i c t e d i n i t s domain t o B; R*A i s t h e image o f A by R ; and R-l*B i s t h e counterimage o f 8 by R * . 9 THEOREM, AXIOMATIC SET THEORY (ix) D R-' 37 = R* D R . The proof i s easy. 10 THEOREM, (i) (R* A ) n 8 = o- A (ii) A n (~-l* 2. 8) (vii) R* ( A % 2 (R* A ) R-'* 8) (viii) R* ( A n R- 8) 0. R* A c -A . 8. C (iv) R* ( A n 8 ) c (R* A ) n (R* 8 ) (v) = U (R* % (R* 8 ) .

ROLAND0 C H U A Q U I 38 P R O O F OF (v). 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.