Lemma: One-loop enumeration, max\ len=2

(i) A connected one-loop graph is 1PI iff every internal line lies on its unique cycle, since a line off the cycle is a bridge. So after the quadratic (${\kappa }^0$, two-field) terms are absorbed into propagators, a one-loop 1PI graph is a cycle of interaction vertices, each with exactly two internal legs. (ii) Every retained interaction vertex has coupling order ${\kappa }^n$ with $n\ge 1$. Mechanically, the ${\kappa }^0$ inventory contains only two-field monomials. (iii) Hence a ${\kappa }^{\le 2}$ contribution has at most two vertices, and max_len=2 is exhaustive. (iv) The auxiliary $b$ occurs only in ${\kappa }^0$ bilinears ($b^2$, $bh$, $b{\bar{c}}^{\ast }$; checked). Its Gaussian elimination produces only bilinear (kinetic or source-bilinear) terms, never an omitted interaction. (v) Source-linear monomials (one quantum leg) cannot sit on the cycle. Attached off it, they create a bridge, i.e.\ a one-particle-reducible graph, so they cannot create extra 1PI coefficients.