“…We will set the Sperner's family as follows: 4, 6, 7, 8, 9, 10}, S 3 = S 6 ∪ S 7 ∪ D 3 = {5, 11, 12, 13, 14, 15}, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}. Process of formation sets S is presented in the figure 2. 2,3,4,5,6,7,8,11,12,13,14, 15}, 2, 3, 4, 5, 6, 7, 8, 9, 10, 14, 15}, 2,3,4,5,6,7,8,9,10,11,12, 13}, k 51 = 0, k 52 = 0, k 53 = 0, k 54 = 0, k 56 = 0, k 57 = 0, k 61 = 0, k 62 = 0, k 63 = 0, k 64 = 0, k 65 = 0, k 67 = 0, k 71 = 0, k 72 = 0, k 73 = 0, k 74 = 0, k 75 = 0, k 76 = 0.…”