s1129 — n–p mass difference: two-sign decomposition with house objects
  world: m_n − m_p = 1.29333 MeV = QCD +2.52 + QED -1.00 (lattice, calibration-grade) · β endpoint 0.78233 MeV

(1) THE QED HALF — Coulomb self-energy of one charged cell (α as declared control)
  uniform ball, proton charge radius 0.8414 fm (calibration-grade): (3/5)αħc/r_p = 1.027 MeV → proton HEAVIER by ≈ 1.03 MeV (lattice QED: 1.00 ± 0.16)
  [PASS] house-style Coulomb self-energy at the measured proton radius reproduces the lattice QED magnitude within 1σ (sign: proton heavier ✓)
  with the COUNT radius instead (α cell, 2.169 fm — the only registered nuclear length): 0.398 MeV — a proton is not an α cell; no registered proton radius exists (the count map has no depth for a single nucleon)

(2) THE QCD HALF — what registered object could carry the u→d cost of ≈ +2.5 MeV?
  required flip energy (to net +1.29333 after the Coulomb −1.03): 2.320 MeV (lattice QCD 2.52 ± 0.17)
  registered constituent masses: m_u^const = m_d^const = Λ_G₂φ^½ (p2) — EQUAL ⇒ the framework's QCD half is ZERO by construction
  [PASS] without an e₇-flip energy the framework predicts the WRONG SIGN for m_n − m_p (Coulomb alone ⇒ proton heavier)
  DECLARED MENU of registered energies vs the required 2.32 MeV (coverage: 9 items, ±5% → ~20% of [0.5, 9] MeV in log):
    p9 intra-cell pair penalty φ⁻¹²√σ/2          0.691 MeV  (-70.2% vs required; -72.6% vs lattice QCD)
    φ⁻³T₃                                        1.708 MeV  (-26.4% vs required; -32.2% vs lattice QCD)
    s1123 calibrated pair coupling c             2.101 MeV  (-9.4% vs required; -16.6% vs lattice QCD)
    m_e·φ³                                       2.165 MeV  (-6.7% vs required; -14.1% vs lattice QCD)
    ½φ⁻¹T₃ (s1123 menu pair coupling)            2.236 MeV  (-3.6% vs required; -11.3% vs lattice QCD)
    3·m_e·φ                                      2.480 MeV  (+6.9% vs required; -1.6% vs lattice QCD)
    φ⁻²T₃                                        2.764 MeV  (+19.1% vs required; +9.7% vs lattice QCD)
    ½T₃                                          3.618 MeV  (+55.9% vs required; +43.6% vs lattice QCD)
    κ·m_N (weak-isospin signature ×½)            8.328 MeV  (+258.9% vs required; +230.5% vs lattice QCD)
  within 5% of the required flip energy: ['½φ⁻¹T₃ (s1123 menu pair coupling)']
  ⇒ the s1123 pair coupling (½φ⁻¹T₃ = 2.236 MeV, or its Ca-calibrated 2.10) sits 4–10% from the required u→d cost. FODDER: it is
    the energy of adding a NEUTRON PAIR per α cell, not of flipping one quark; a menu member, not a derivation.

(3) WHAT THE X-POST CLAIM WOULD HAVE TO SHOW (the same bar we hold ourselves to)
  ΔLk = ±1 ↦ 0.782 MeV is ONE integer label on ONE world number. The lattice decomposition is TWO numbers of OPPOSITE sign; a
  linking-number model must produce (a) a Coulomb-like −1.0 MeV that does not depend on Lk, and (b) a +2.5 MeV that does, and
  say why the electron's 0.511 sits where it does. Our own machinery reproduces (a) trivially (with α loaded) and CANNOT produce
  (b): the u↔d object is a ℤ₂ without an energy (s1124). That is the honest state of BOTH the post's model and ours on this number.

VERDICT
  Scoped NO-GO with the obstruction named: the framework has the QED half (Coulomb, α external) and no QCD half — the e₇
  conjugation carries no registered energy, and the constituent u/d masses are equal by construction. Without a flip energy the
  predicted sign of m_n − m_p is WRONG. Candidate object for the lane: the same non-scalar Hamiltonian A1545 named (it must act on
  the doublet to price u→d), with the s1123 pair coupling as the number to beat. m_n − m_p joins the ledger as a NEW UNPRICED ROW
  (it appears nowhere in the suite: 0 tex hits for 1.293 / 0.782). The X post is a foil, not a lead.

s1129: ALL CHECKS PASS
