“…Observe that by Step 2, both f 2 and δh 3 vanish on 5-tuples z satisfying n 1 (z) + n 2 (z) 7, so that the same holds for f 3 . We will now prove, step by step, that f 3 also vanishes on 5-tuples z with (n 1 (z), n 2 (z)) = (3, 5), (4,4), (4,5), and (5,5). In all but one subcase, the strategy is the same as in most of the proof of Step 2.…”