Provedalphabaguettenoether-checksumattention

Checksum C1: summed over all key positions, the key gradient is exactly zero

For all $\htmlData{mx=1}{T, D \in \htmlData{mx=0}{\mathbb{N}}}$ , $\htmlData{mx=9}{P, \mathit{dP} : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{T}} \mathrel{\htmlData{mx=8}{\to}} \htmlData{mx=5}{\htmlData{mx=3}{\operatorname{Fin}}\,\htmlData{mx=4}{T}} \mathrel{\htmlData{mx=7}{\to}} \htmlData{mx=6}{\mathbb{R}}}$ , $\htmlData{mx=9}{q : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{T}} \mathrel{\htmlData{mx=8}{\to}} \htmlData{mx=5}{\htmlData{mx=3}{\operatorname{Fin}}\,\htmlData{mx=4}{D}} \mathrel{\htmlData{mx=7}{\to}} \htmlData{mx=6}{\mathbb{R}}}$ , $\htmlData{mx=1}{\mathit{scale} \in \htmlData{mx=0}{\mathbb{R}}}$ and $\htmlData{mx=3}{c \in \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{D}}}$ , if $\htmlData{mx=6}{\forall i,\ \htmlData{mx=4}{\sum_{j} \htmlData{mx=3}{\htmlData{mx=0}{P}(\htmlData{mx=1}{i}, \htmlData{mx=2}{j})}} \mathrel{\htmlData{mx=5}{=}} 1}$ , then

\begin{aligned} & \htmlData{mx=29}{\sum_{j} \htmlData{mx=28}{\htmlData{mx=1}{\mathit{scale}} \cdot \htmlData{mx=27}{\sum_{i} \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)
∑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)

theorem nc-key-gradient-sums-to-zero {T D : ℕ} (P dP : Fin T → Fin T → ℝ) (q : Fin T → Fin D → ℝ) (scale : ℝ) (hP : ∀ i, ∑ j, P i j = 1) (c : Fin D) :
    ∑ j, scale * ∑ i, (P i j * (dP i j - ∑ k, P i k * dP i k)) * q i c = 0

With dKⱼ = scale·Σᵢ Rᵢⱼ·qᵢ and each row of R summing to zero, Σⱼ dKⱼ = 0 in every component, for any queries, any probabilities with unit row sums (causal or not), and any upstream gradient. One reduction per head per layer checks it. In the bug-injection experiment it flags Baguette's wrong-V-layout, transpose-overwrite and store-race bugs at 186–1,060× the clean BF16 residual. Part of 'Noether checksums for Baguette's backward kernels'.

How it was proved

Proved: 3 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: 23 lines of Lean

Checksum C1: summed over all key positions, the key gradient is exactly zero

How it was proved

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