e_022103
plain-language theorem explainer
For the six Fin-4 indices (0,2,2,1,0,3), the folded coupling numerator equals eight times the explicit integer table entry. Gravity analysts certifying the 4D Regge midpoint M2 TT identity cite this as one of 256 kernel-decided index cases. The proof is a single decide on two concrete integers.
Claim. With six indices in $\mathrm{Fin}\,4$ fixed to $(0,2,2,1,0,3)$, the coupling-fold numerator equals eight times the explicit closed-form integer table: $N_2(0,2,2,1,0,3)=8\,Z(0,2,2,1,0,3)$.
background
In the 4D Regge midpoint M2 TT analysis, two integer maps on six $\mathrm{Fin},4$ indices are compared. The numerator $N_2$ (Lean m2Num) folds a fixed coupling list, adding one contribution per term. The companion map $Z$ (Lean explicitZ) is a sparse pattern-matched table of small integers (entries such as $4$, $-2$, and so on).
This module is chunk 2 of a 256-case kernel certification that $N_2=8Z$ pointwise. The local setting is pure finite enumeration: each chunk lemma pins one concrete sextuple so the assembly theorem can exhaust $\mathrm{Fin},4^6$ by fin_cases.
proof idea
Both sides are closed integer terms once the six indices are literals. The proof is the one-tactic decide: the kernel normalizes the fold on the left and the table lookup on the right, then checks integer equality. No lemmas beyond the two definitions are invoked.
why it matters
This case is consumed by m2Num_eq_eight_explicitZ, which asserts $N_2(a,b,c,d,i,j)=8Z(a,b,c,d,i,j)$ for every sextuple in $\mathrm{Fin},4$. That global factor-of-eight identity is a numerator step inside the Regge exact midpoint M2 TT certification for 4D gravity. Chunking into 256 decide goals keeps each kernel check tiny while the assemble theorem glues the cover. It does not itself touch continuum limits, phi-ladder mass formulas, or the T0–T8 forcing chain.
Switch to Lean above to see the machine-checked source, dependencies, and usage graph.