e_012330
plain-language theorem explainer
For the six-index slot (0,1,2,3,3,0) on Fin 4, the folded numerator m2Num equals eight times the closed-form kernel value explicitZ. Gravity analysts cite it when assembling the exact midpoint M2TT identity in 4D Regge calculus. The proof is a single kernel decide on concrete integers.
Claim. For indices $(a,b,c,d,i,j)=(0,1,2,3,3,0)$ in $\mathrm{Fin}\,4$, the folded coupling numerator satisfies $m_2^{\mathrm{num}}(0,1,2,3,3,0)=8\,Z_{\mathrm{explicit}}(0,1,2,3,3,0)$.
background
The ambient module certifies, one index tuple at a time, that the 4D Regge midpoint numerator m2Num is identically eight times a sparse closed-form table explicitZ. Both maps take six Fin 4 arguments (four simplex labels and a pair of edge indices) and return an Int.
m2Num is defined by folding a fixed coupling list: each term contributes an integer weight via contrib, and the fold starts at 0. explicitZ is the matching piecewise table (nonzero only on a short list of index patterns such as $(0,0,1,1,2,2)\mapsto 4$ and $(0,0,1,2,1,2)\mapsto -2$). The module doc frames the work as chunk 1 of 256 kernel decides establishing $m_2^{\mathrm{num}}=8\cdot Z_{\mathrm{explicit}}$.
proof idea
One-line computational proof: by decide. Lean evaluates both sides on the concrete six-tuple $(0,1,2,3,3,0)$ inside the integer kernel and checks equality. No lemmas are invoked beyond the definitions of m2Num and explicitZ.
why it matters
This is one atomic cell in the exhaustive grid that feeds m2Num_eq_eight_explicitZ in the assemble module. That parent theorem states $\forall a,b,c,d,i,j:\mathrm{Fin},4,; m_2^{\mathrm{num}}=8,Z_{\mathrm{explicit}}$ and discharges the quantifiers by nested fin_cases. The identity is the algebraic core of the exact midpoint M2TT certificate for 4D Regge gravity in the Recognition Science stack: once numerator and closed form agree everywhere, curvature and mass-coupling bookkeeping can quote a single sparse table instead of a fold. It does not itself touch the T0–T8 forcing chain, but it hardens the gravity-side numerics that sit downstream of the eight-tick and $D=3$ landmarks.
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