B_pow_sum_eq_A

Open Problem O1 asks why each particle sector receives a specific binary power $B_{\mathrm{pow}}$ and rung offset $r_0$ from the counting layer. The Yardstick Assignment Principle answers by coupling each sector to a level of the 3-cube hierarchy: leptons to passive edges, up quarks and electroweak to the active edge, down quarks to total edges.

$B_{\mathrm{pow}}$ is defined sector-wise from cube edge counts: lepton $-(2 E_{\mathrm{passive}}) = -22$, up $-A = -1$, down $2 E_{\mathrm{total}} - 1 = 23$, electroweak $+A = +1$. Here $A$ is the active-edge count per tick (equal to 1), $E_{\mathrm{passive}} = 11$, and $E_{\mathrm{total}} = 12$. The sibling identity $B_{\mathrm{pow}}$-sum already shows these four integers add to 1; the present statement rewrites that total as $A$ in counting-layer form.

Two-step calc. First apply the sibling lemma B_pow_sum, which evaluates the four-sector sum to the integer 1 from the closed formulas for each $B_{\mathrm{pow}}$. Second, native_decide discharges $1 = (A : \mathbb{Z})$ using the definition $A = 1$. No further algebraic rewriting is needed.

This is the counting-layer form of the sector-power partition that underwrites O1. The module records the structural observation that the $B_{\mathrm{pow}}$ values split into complementary pairs whose magnitudes track how much each sector borrows from the cube edge network (leptons borrow $2 E_{\mathrm{passive}}$, quarks and EW sit on the active edge $A$). Stating the sum as equality with $A$ rather than bare $1$ ties the identity to the same active-edge constant that appears in the gap derivation ($\eta_B \cdot \Theta_{\mathrm{crit}} = \varphi^A$) and in the mass-anchor yardstick. No downstream theorems currently depend on it; it is a verification checkpoint that the four sector exponents are a partition of the active-edge unit, not an ad-hoc assignment.