Full proofs of lemmas Lemma 3 Suppose length-n vectors x and y differ at exactly k values, and Summary: Full proofs of lemmas Lemma 3 Suppose length-n vectors x and y differ at exactly k values, and for these values yi = xi + , where is a positive constant. Denote w = min{k, m+ - 1}. Then, the following inequality holds: L(x) L(y) L(x) + · w. (18) Proof. We will define a candidate sum for x as any sum of m+ - 1 distinct l(i, xi, p) values as in (3). By (4), L(x) is the largest candidate sum for x. First, we prove L(x) L(y). Consider the candidate sum S for y com- puted by selecting the same i and p values as in L(x). Because for all i, xi yi, L(x) L(y). Because L(y) must be the largest candidate sum for y, S L(y). Therefore, L(x) L(y). Next, we prove L(y) L(x) + · w by contradiction. Suppose L(y) > L(x) + · w. Consider the candidate sum T for x computed by selecting the same i and p values as in L(y). Observe that at most w terms contribute to the difference between L(y) and T. When two such terms differ, we have xi = yi - (xi = yi otherwise). Thus, T L(y)- ·w, and hence, T > L(x), which contradicts the fact that L(x) is a maximal candidate sum for x. Lemma 4 If y is compliant and there is a j such that yj > (L(y) + S() + U()Di - Ci)/m, then there exists a strictly smaller vector x that is also Collections: Computer Technologies and Information Sciences