Provedalphabaguettenoether-checksumcontrast

Exact detection power of C2: its residual equals the total corruption of the dO copy the kernel used

For all $\htmlData{mx=1}{T, D \in \htmlData{mx=0}{\mathbb{N}}}$ , $\htmlData{mx=9}{P : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{T}} \mathrel{\htmlData{mx=8}{\to}} \htmlData{mx=5}{\htmlData{mx=3}{\operatorname{Fin}}\,\htmlData{mx=4}{T}} \mathrel{\htmlData{mx=7}{\to}} \htmlData{mx=6}{\mathbb{R}}}$ , $\htmlData{mx=9}{\mathit{dO}, \mathit{dO}' : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{T}} \mathrel{\htmlData{mx=8}{\to}} \htmlData{mx=5}{\htmlData{mx=3}{\operatorname{Fin}}\,\htmlData{mx=4}{D}} \mathrel{\htmlData{mx=7}{\to}} \htmlData{mx=6}{\mathbb{R}}}$ 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

\begin{aligned} & \htmlData{mx=17}{\htmlData{mx=11}{\sum_{j} \htmlData{mx=10}{\sum_{i} \htmlData{mx=9}{\htmlData{mx=4}{\htmlData{mx=1}{P}(\htmlData{mx=2}{i}, \htmlData{mx=3}{j})} \cdot \htmlData{mx=8}{\htmlData{mx=5}{\mathit{dO}'}(\htmlData{mx=6}{i}, \htmlData{mx=7}{c})}}}} - \htmlData{mx=16}{\sum_{i} \htmlData{mx=15}{\htmlData{mx=12}{\mathit{dO}}(\htmlData{mx=13}{i}, \htmlData{mx=14}{c})}}} \\ & \quad \htmlData{mx=0}{=} \htmlData{mx=27}{\sum_{i} (\htmlData{mx=26}{\htmlData{mx=21}{\htmlData{mx=18}{\mathit{dO}'}(\htmlData{mx=19}{i}, \htmlData{mx=20}{c})} - \htmlData{mx=25}{\htmlData{mx=22}{\mathit{dO}}(\htmlData{mx=23}{i}, \htmlData{mx=24}{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: dO, dO' range over Fin T → Fin D → ℝ: the statement must hold for every choice. ((dO dO' : Fin T → Fin D → ℝ))
∀For all: The statement after the comma holds for every i. (∀ i, ∑ j, P i j = 1)
-Subtraction: ∑ j, ∑ i, P(i, j) · dO'(i, c) minus ∑ i, dO(i, c). On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. (∑ j, ∑ i, P i j * dO' i c - ∑ i, dO i c)
∑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-checksum-detection-power {T D : ℕ} (P : Fin T → Fin T → ℝ) (dO 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 = ∑ i, (dO' i c - dO i c)

If the key kernel computes dV from a corrupted copy dO′ (as in Baguette's packed-transpose bug), the C2 residual Σⱼ dV′ⱼ − Σᵢ dOᵢ equals Σᵢ (dO′ᵢ − dOᵢ). So C2 detects exactly the corruptions that change the column sum, and nothing about P is needed. 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: 16 lines of Lean

Exact detection power of C2: its residual equals the total corruption of the dO copy the kernel used

How it was proved

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