e_202200
plain-language theorem explainer
At multi-index (2,0,2,2,0,0) the folded Regge coupling numerator equals eight times the explicit integer kernel. Gravity analysts assembling the 4D midpoint M2 TT identity cite this as one finite-case certificate among the 256-way split. The proof is a single kernel decision of integer equality.
Claim. For indices $a=2$, $b=0$, $c=2$, $d=2$, $i=0$, $j=0$ in $\mathrm{Fin}\,4$, the folded coupling numerator $m_2^{\mathrm{num}}(a,b,c,d,i,j)$ equals $8$ times the explicit integer kernel value $Z(a,b,c,d,i,j)$.
background
In the 4D Regge analysis of the exact midpoint M2 TT identity, two integer kernels are compared pointwise on $(\mathrm{Fin},4)^6$. The numerator $m_2^{\mathrm{num}}$ is the fold of a fixed coupling list: it sums a contribution function over that list at the given six indices. The companion explicit kernel $Z$ is a sparse pattern match returning small integers (typically $\pm 2$, $\pm 4$, or $0$) on its support.
This module is chunk 8 of a 256-way case split. Each local theorem asserts $m_2^{\mathrm{num}}=8\cdot Z$ at one concrete multi-index. The global factor of $8$ is the normalization that makes the identity hold everywhere once all cases are assembled.
proof idea
With all six $\mathrm{Fin},4$ arguments fixed, both sides reduce to concrete integers. The proof is the one-line tactic decide, which checks integer equality in the kernel. No algebraic lemmas or rewriting are used.
why it matters
The assembly theorem m2Num_eq_eight_explicitZ quantifies over all six indices by exhaustive fin_cases and consumes each chunk certificate, including this one, to obtain
$\forall a,b,c,d,i,j,; m_2^{\mathrm{num}}(a,b,c,d,i,j)=8,Z(a,b,c,d,i,j)$.
Closing the pointwise identity is a necessary algebraic step toward the exact midpoint M2 TT identity in 4D Regge gravity inside the Recognition Science gravity stack. It does not itself touch the forcing chain (T0–T8) or the Recognition Composition Law; it is pure discrete-kernel bookkeeping.
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