 
Summary: Full proofs of lemmas
Lemma 3 Suppose lengthn 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
