proofs
Provedalphabaguettenoether-checksumreview-2

Honest blind spot: C1 still reads zero if the key kernel skips ANY subset of query rows (truncated grid, missing head)

For all T,D∈N\htmlData{mx=1}{T, D \in \htmlData{mx=0}{\mathbb{N}}}, P,dP:Fin⁡ T→Fin⁡ T→R\htmlData{mx=9}{P, \mathit{dP} : \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}{T}} \mathrel{\htmlData{mx=7}{\to}} \htmlData{mx=6}{\mathbb{R}}}, q:Fin⁡ T→Fin⁡ D→R\htmlData{mx=9}{q : \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}}}, S∈Finset⁡ Fin⁡ T\htmlData{mx=5}{S \in \htmlData{mx=4}{\htmlData{mx=0}{\operatorname{Finset}}\,\htmlData{mx=3}{\htmlData{mx=1}{\operatorname{Fin}}\,\htmlData{mx=2}{T}}}} and c∈Fin⁡ D\htmlData{mx=3}{c \in \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{D}}}, if ∀i, ∑jP(i,j)=1\htmlData{mx=6}{\forall i,\ \htmlData{mx=4}{\sum_{j} \htmlData{mx=3}{\htmlData{mx=0}{P}(\htmlData{mx=1}{i}, \htmlData{mx=2}{j})}} \mathrel{\htmlData{mx=5}{=}} 1}, then

∑j∑i∈SP(i,j)⋅(dP(i,j)−∑kP(i,k)⋅dP(i,k))⋅q(i,c)=0\begin{aligned} & \htmlData{mx=28}{\sum_{j} \htmlData{mx=27}{\sum_{i \in \htmlData{mx=1}{S}} \htmlData{mx=26}{\htmlData{mx=21}{\htmlData{mx=5}{\htmlData{mx=2}{P}(\htmlData{mx=3}{i}, \htmlData{mx=4}{j})} \cdot \left(\htmlData{mx=20}{\htmlData{mx=9}{\htmlData{mx=6}{\mathit{dP}}(\htmlData{mx=7}{i}, \htmlData{mx=8}{j})} - \htmlData{mx=19}{\sum_{k} \htmlData{mx=18}{\htmlData{mx=13}{\htmlData{mx=10}{P}(\htmlData{mx=11}{i}, \htmlData{mx=12}{k})} \cdot \htmlData{mx=17}{\htmlData{mx=14}{\mathit{dP}}(\htmlData{mx=15}{i}, \htmlData{mx=16}{k})}}}}\right)} \cdot \htmlData{mx=25}{\htmlData{mx=22}{q}(\htmlData{mx=23}{i}, \htmlData{mx=24}{c})}}}} \\ & \quad \htmlData{mx=0}{=} 0 \end{aligned}
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
P, dP range over Fin T → Fin T → ℝ: the statement must hold for every choice. ((P dP : Fin T → Fin T → ℝ))
∀For all
The statement after the comma holds for every i. (∀ i, ∑ j, P i j = 1)
Finset
Finset: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (Finset)
∑Summation
Add up P(i, j) for every j. (∑ j, P i j)
=Equality
∑ j, P(i, j) and 1 are the same. (∑ j, P i j = 1)
⟹Implies
If Fin(T), then ℝ. (Between types, A → B is the type of functions from A to B.) (Fin T → ℝ)
RReal numbers
The real numbers: every point on the number line. (ℝ)
-Subtraction
dP(i, j) minus ∑ k, P(i, k) · dP(i, k). On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. (dP i j - ∑ k, P i k * dP i k)
·Multiplication
P(i, k) times dP(i, k). (P i k * dP i k)

Each query row of the score gradient sums to zero on its own, so dropping any subset S of query rows (a skipped tile, a truncated grid, an entire GQA head) leaves Σⱼ dKⱼ = 0. The experiment confirms C1 misses B7 (skipped tile) and B8 (missing head), while C2 and C3 catch both. 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
19 lines of Lean
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