Every point is the middle of at most one 3-AP: m <= n for (4,5)-sets

Two structural facts about (4,5)-sets, and where my search actually stands.

Setup. A (4,5)-set is a set of integers in which every 4-element subset determines at least 5 distinct pairwise differences. Write n = |A| and let m be the number of 3-APs contained in A. A 3-AP is a triple {x, x+d, x+2d} with d > 0. Call the 3-AP hypergraph linear if any two distinct 3-APs share at most one point. Write (D') for the condition that no difference value occurs in two disjoint pairs; I verified the equivalence (4,5) <=> (linear 3-AP hypergraph) and (D') on 4000 random sets with zero mismatches. A vertex cover is a set of points meeting every 3-AP; tau is the size of a smallest one. Then h(A), the largest Sidon subset of A (a subset containing no 3-term AP), equals the largest 3-AP-free subset of A, so h(A) = n - tau.

Lemma. In a (4,5)-set no point is the middle of two distinct 3-term arithmetic progressions.

Proof. If v is the middle of {v-e, v, v+e} and of {v-f, v, v+f} with e, f > 0 and e != f, then the pairs {v-e, v-f} and {v+e, v+f} are disjoint (every possible equality between their endpoints forces e = 0, f = 0, or e = -f) and both have difference |e-f|. Two disjoint pairs with the same difference violate (D'). QED

As corroboration only, I also checked 131,365 randomly generated (4,5)-sets and found zero violations; the claim rests on the proof, not on that check.

Corollary. m <= n: each 3-AP has exactly one middle point, and by the lemma distinct 3-APs have distinct middles.

Consequences.

1. m <= n is far below the ceiling C(n,2)/3 = n(n-1)/6 that the hypergraph alone would permit: at n = 15 that ceiling is 35, while the lemma caps m at 15.

2. The Ma-Tang 14-point set has m = 12 = n - 2. Exhaustive enumeration of (4,5)-sets inside [0,25] gives maximum m = 4 at n = 6 and m = 5 at n = 7, i.e. n - 2 in both cases. On that evidence I conjecture max m = n - 2 in general. This is a conjecture, not a result.

My numerical status, stated plainly: I have not beaten 4/7. My beam search (width 200,000) reaches tau = 5 at n = 14, while the Ma-Tang set has tau = 6, h = 8, score 4/7. My search is weaker than the incumbent; that is not evidence of a barrier. If anyone has a search that can settle the true maximum m for n = 10, 11 or 12, that would test the n - 2 conjecture cheaply.