Provedalphabaguettenoether-checksumreview-2

If the randomized pairing passes for every M, the dK error is invisible to all keys: Σⱼ eⱼ ⊗ kⱼ = 0

For all $\htmlData{mx=1}{T, D \in \htmlData{mx=0}{\mathbb{N}}}$ , $\htmlData{mx=9}{e, k : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{T}} \mathrel{\htmlData{mx=8}{\to}} \htmlData{mx=5}{\htmlData{mx=3}{\operatorname{Fin}}\,\htmlData{mx=4}{D}} \mathrel{\htmlData{mx=7}{\to}} \htmlData{mx=6}{\mathbb{R}}}$ and $\htmlData{mx=3}{a_{0}, b_{0} \in \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{D}}}$ , if $\htmlData{mx=27}{\forall M : \htmlData{mx=20}{\htmlData{mx=18}{\operatorname{Fin}}\,\htmlData{mx=19}{D}} \mathrel{\htmlData{mx=26}{\to}} \htmlData{mx=23}{\htmlData{mx=21}{\operatorname{Fin}}\,\htmlData{mx=22}{D}} \mathrel{\htmlData{mx=25}{\to}} \htmlData{mx=24}{\mathbb{R}},\ \htmlData{mx=16}{\sum_{j} \htmlData{mx=15}{\sum_{a} \htmlData{mx=14}{\htmlData{mx=3}{\htmlData{mx=0}{e}(\htmlData{mx=1}{j}, \htmlData{mx=2}{a})} \cdot \left(\htmlData{mx=13}{\sum_{b} \htmlData{mx=12}{\htmlData{mx=7}{\htmlData{mx=4}{M}(\htmlData{mx=5}{b}, \htmlData{mx=6}{a})} \cdot \htmlData{mx=11}{\htmlData{mx=8}{k}(\htmlData{mx=9}{j}, \htmlData{mx=10}{b})}}}\right)}}} \mathrel{\htmlData{mx=17}{=}} 0}$ , then

\htmlData{mx=9}{\sum_{j} \htmlData{mx=8}{\htmlData{mx=3}{\htmlData{mx=0}{e}(\htmlData{mx=1}{j}, \htmlData{mx=2}{a_{0}})} \cdot \htmlData{mx=7}{\htmlData{mx=4}{k}(\htmlData{mx=5}{j}, \htmlData{mx=6}{b_{0}})}}} \mathrel{\htmlData{mx=10}{=}} 0

What the notation means · or point at any symbol

NNatural numbers: The natural numbers 0, 1, 2, …. In Lean they start at 0. (ℕ)
x ∈ TVariables: T, D range over natural numbers: the statement must hold for every choice. ((T D : ℕ))
Fin: Fin: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (Fin)
f(x)Function application: Fin applied to T. Lean writes f x; papers write f(x). (Fin T)
x : TVariables: e, k range over Fin T → Fin D → ℝ: the statement must hold for every choice. ((e k : Fin T → Fin D → ℝ))
∀For all: The statement after the comma holds for every M. (∀ M : Fin D → Fin D → ℝ, ∑ j, ∑ a, e j …)
⟹Implies: If Fin(D), then ℝ. (Between types, A → B is the type of functions from A to B.) (Fin D → ℝ)
RReal numbers: The real numbers: every point on the number line. (ℝ)
=Equality: ∑ j, ∑ a, e(j, a) · (∑ b, M(b, a) · k(j, b)) and 0 are the same. (∑ j, ∑ a, e j a * (∑ b, M b a * k j b) …)
∑Summation: Add up M(b, a) · k(j, b) for every b. (∑ b, M b a * k j b)
·Multiplication: M(b, a) times k(j, b). (M b a * k j b)

theorem nc2-pairing-family-detection {T D : ℕ} (e k : Fin T → Fin D → ℝ) (h : ∀ M : Fin D → Fin D → ℝ, ∑ j, ∑ a, e j a * (∑ b, M b a * k j b) = 0) (a₀ b₀ : Fin D) :
    ∑ j, e j a₀ * k j b₀ = 0

Detection power of the pairing family. If a computed dK′ = dK + e passes Σⱼ⟨eⱼ, Mᵀkⱼ⟩ = 0 for every matrix M, then every entry of the D×D matrix Σⱼ eⱼ kⱼᵀ vanishes (take M = an elementary matrix). Escaping errors must be orthogonal, in this tensor sense, to the keys; permuted rows (B4) generically are not. Part of 'Noether checksums for Baguette's backward kernels' (review pass 2).

How it was proved

Proved: 4 October 2026
By: Ajax, with the agent “Baguette math & proofs (Claude Code, 2026-10-03)”
Checked by: Lean's kernel (Lean 4.33.1, with Mathlib); the proof is signed and replays offline
Axioms: Classical.choice, Quot.sound, propext
Proof: 13 lines of Lean

If the randomized pairing passes for every M, the dK error is invisible to all keys: Σⱼ eⱼ ⊗ kⱼ = 0

How it was proved

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