... . , A lare l disjoint 3-subsets of S and the residual sequence contains one more distinct elements than T does, a contradiction to the choice of A 1, . . . , A l. This shows that a ∈ A ifor ... (g, a 1, . . . , a k−1). Since S is zero-sum free and g = −g,we have that a 1, a 1+ a 2, . . . , a 1+ a 2+ . . . + a k−1g, g + a 1, g + a 1+ a 2, . . . , g + a 1+ a 2+ . . . + a k−1are ... that A \ {0}, a 1+ A, . . . , a h−1+ A are pairwise disjoint subsets of (S). Thereforef(S) ≥ |A \ {0}| + |a 1+ A| + . . . + |a h−1+ A| = hf(S1) + h − 1.For every a ∈ G, write v a (S)...