Problem Set 12: (a) Let X, Y, and Z be arbitrary sets. Use an element argument to prove that X n (Y -Z) =(XnY) -(XnZ). (b) Let A and B be arbitrary sets. Use an element argument to prove that A U (B - A) = AUB.Supposition etxe xn ( y- 7 ) be an arbitrary element . Cool : x(y-2) = (xny ) - (x02 ) Deduction : X 6 X 1 ( 7 - 2 ) 1 = intersection , if AMB that means ME A and B both ' - ' means if ke A-B that means KEA such that Keb now , if re XA( y - 2 ) .( and is used for intersection ) = x belongs to both * & (x- 2) - x ex and x6 ( y -2)- now open this using A- B - xEX and ( x e y such that no z ) . explained above - ( *EX and KEY) such that * $7 ( now use associativityel brackets) 3 x E(x and y both ) such that 4 2 > xe(xny ) such that nez ], This is the same -sign property 2 xe ( xny ) - 7 3 bie. came as ( xny ) - z now cansides BAB , ANBSA & AnBCB alwaysAUD ( A - B ) ARB (B - A ) A B Clearly A UCB - A) = AUB -@ &To prove : KB we show Xep & BEX . Let us show KOB ine. Auce-me ADB To show XSB we take an element in a then we show it belongstop Supposition : Let me Au(e - A ) be an aobituary element . local : AU(B - A) C AUB Deduction: s KEA ox X E B -A " ( Union means ' or " ) & AUD [ . . ACAUB ] Case I. if KE B-A - KEB such that he ASKED = YEAUB ['B SAUB] Both conditions imply * . YEA . x Xe B- A =) HEAUB 5. AU(B - A ) & AUB - Condosion . Dal . AUB CAULD -A ) Supposition i now let ME AUB Deduction: 2 * GA of XEB Cos e I ! ( * E A = ) x e AU ( B - A ) [AC AUX, WhenP X = B-A ) Is we can take any set X & : will still hold . May lose It . In WEB J x belong to ( B -A) as it condoms ( B-A) all elements of A that are not in A ox x may belong to AAB Art AND 5 ( AnB)U( B - A) HE (AAB ) U( B - A ) > KE AU(B-A) Jas ANDSA'] In both cases HE AU( B - A ) AUB C AU(B - A ) from Of a - AUB = AU(B - A )It means if KE AnB ) neA also & WEB also. . AlB is always a subset of A and B both. It any element doesnot belong to ~ A then, it won't belong to AMB also . We use the same property in Z If KEZ That means x * ( XOZ ) also . so replace 2 by ( xAZ ) we get . 3 x6 (xay ) - (xnz ) . we started from nexn( y -z ) and got me(xny ) - (onz ) - 2( E X 1 ( - 2 ) ) HE ( xay ) - (XAZ ) ( we know that if ve A > XEB that means A CB ) xn ( y -2) = (xny ) - (x1z ) conclusion]