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2. Let U be a set, and let A CU. Recall the indicator function XA: U → Z2 defined by XA() : S 1, XEA 10, x¢ A. Now, let A, B(b) Prove that Vx € U, XAAB(x) = x1(x) + XB(x), where addition is taken modulo 2 (so that 1+1 = 0).

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27 Here, А, Зco , 4 Адре (А–В) (В - А) if ite АДВ, ече (А -В) or ce (3-4) re, m A4 тф в од че в д и фn clearly nine A & & &case, For any Then XA AB ar & EB & & & A Them X B (A) = 1 & XA (2) = 0 Then XA (0) + Xa ( ): 01 | | i., XAAB (A) XA (x) + xp

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