“…Then y 1 → (1, 1, 0, 1, 1, 1), y 2 → (2, 0, 1, 1, 2, 2), y 3 → (0, 2, 2, 2, 1, 2), and the obtained three elements generate S = 3 6 L as an SBIA. Note that S can not be generated by three generators as an SBA, but at least by four generators, see [11].…”