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Checksum C3: Σᵢ⟨dQᵢ, qᵢ⟩ = Σⱼ⟨dKⱼ, kⱼ⟩, tying the query and key backward kernels together

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

\begin{aligned} & \htmlData{mx=19}{\sum_{i} \htmlData{mx=18}{\sum_{c} \htmlData{mx=17}{\htmlData{mx=12}{\htmlData{mx=1}{\mathit{scale}} \cdot \htmlData{mx=11}{\sum_{j} \htmlData{mx=10}{\htmlData{mx=5}{\htmlData{mx=2}{R}(\htmlData{mx=3}{i}, \htmlData{mx=4}{j})} \cdot \htmlData{mx=9}{\htmlData{mx=6}{k}(\htmlData{mx=7}{j}, \htmlData{mx=8}{c})}}}} \cdot \htmlData{mx=16}{\htmlData{mx=13}{q}(\htmlData{mx=14}{i}, \htmlData{mx=15}{c})}}}} \\ & \quad \htmlData{mx=0}{=} \htmlData{mx=38}{\sum_{j} \htmlData{mx=37}{\sum_{c} \htmlData{mx=36}{\htmlData{mx=31}{\htmlData{mx=20}{\mathit{scale}} \cdot \htmlData{mx=30}{\sum_{i} \htmlData{mx=29}{\htmlData{mx=24}{\htmlData{mx=21}{R}(\htmlData{mx=22}{i}, \htmlData{mx=23}{j})} \cdot \htmlData{mx=28}{\htmlData{mx=25}{q}(\htmlData{mx=26}{i}, \htmlData{mx=27}{c})}}}} \cdot \htmlData{mx=35}{\htmlData{mx=32}{k}(\htmlData{mx=33}{j}, \htmlData{mx=34}{c})}}}} \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: q, k range over Fin T → Fin D → ℝ: the statement must hold for every choice. ((q k : Fin T → Fin D → ℝ))
=Equality: ∑ i, ∑ c, scale · (∑ j, R(i, j) · k(j, c)) · q(i, c) and ∑ j, ∑ c, scale · (∑ i, R(i, j) · q(i, c)) · k(j, c) are the same. (∑ i, ∑ c, (scale * ∑ j, R i j * k j c) …)
⟹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)
∑Summation: Add up R(i, j) · k(j, c) for every j. (∑ j, R i j * k j c)
·Multiplication: R(i, j) times k(j, c). (R i j * k j c)

theorem nc-query-key-pairing {T D : ℕ} (R : Fin T → Fin T → ℝ) (q k : Fin T → Fin D → ℝ) (scale : ℝ) :
    ∑ i, ∑ c, (scale * ∑ j, R i j * k j c) * q i c = ∑ j, ∑ c, (scale * ∑ i, R i j * q i c) * k j c

With dQᵢ = scale·Σⱼ Rᵢⱼ kⱼ and dKⱼ = scale·Σᵢ Rᵢⱼ qᵢ for the SAME score gradient R, both scalars equal scale·Σᵢⱼ Rᵢⱼ⟨qᵢ, kⱼ⟩. Holds for any R. It catches bugs that preserve sums, such as dK rows stored to permuted addresses (missed by C1, flagged at 378× in the experiment). 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: 24 lines of Lean

Checksum C3: Σᵢ⟨dQᵢ, qᵢ⟩ = Σⱼ⟨dKⱼ, kⱼ⟩, tying the query and key backward kernels together

How it was proved

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