Provedalphabaguettenoether-checksumattention

Checksum C2: Σⱼ dVⱼ = Σᵢ dOᵢ

For all $\htmlData{mx=1}{T, D \in \htmlData{mx=0}{\mathbb{N}}}$ , $\htmlData{mx=9}{P : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{T}} \mathrel{\htmlData{mx=8}{\to}} \htmlData{mx=5}{\htmlData{mx=3}{\operatorname{Fin}}\,\htmlData{mx=4}{T}} \mathrel{\htmlData{mx=7}{\to}} \htmlData{mx=6}{\mathbb{R}}}$ , $\htmlData{mx=9}{\mathit{dO} : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{T}} \mathrel{\htmlData{mx=8}{\to}} \htmlData{mx=5}{\htmlData{mx=3}{\operatorname{Fin}}\,\htmlData{mx=4}{D}} \mathrel{\htmlData{mx=7}{\to}} \htmlData{mx=6}{\mathbb{R}}}$ and $\htmlData{mx=3}{c \in \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{D}}}$ , 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

\htmlData{mx=10}{\sum_{j} \htmlData{mx=9}{\sum_{i} \htmlData{mx=8}{\htmlData{mx=3}{\htmlData{mx=0}{P}(\htmlData{mx=1}{i}, \htmlData{mx=2}{j})} \cdot \htmlData{mx=7}{\htmlData{mx=4}{\mathit{dO}}(\htmlData{mx=5}{i}, \htmlData{mx=6}{c})}}}} \mathrel{\htmlData{mx=16}{=}} \htmlData{mx=15}{\sum_{i} \htmlData{mx=14}{\htmlData{mx=11}{\mathit{dO}}(\htmlData{mx=12}{i}, \htmlData{mx=13}{c})}}

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)
∀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)
·Multiplication: P(i, j) times dO(i, c). (P i j * dO i c)
RReal numbers: The real numbers: every point on the number line. (ℝ)

theorem nc-value-gradient-sum-equals-output-gradient-sum {T D : ℕ} (P : Fin T → Fin T → ℝ) (dO : Fin T → Fin D → ℝ) (hP : ∀ i, ∑ j, P i j = 1) (c : Fin D) :
    ∑ j, ∑ i, P i j * dO i c = ∑ i, dO i c

With dVⱼ = Σᵢ Pᵢⱼ·dOᵢ and unit row sums, the key-side value gradient summed over keys equals the upstream output gradient summed over queries. That is one reduction on each side. In the experiment it is the only checksum that isolates a corrupted dO copy inside the key-backward kernel (Baguette's packed-transpose bug). 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: 14 lines of Lean

Checksum C2: Σⱼ dVⱼ = Σᵢ dOᵢ

How it was proved

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