Provedalphabaguettealgorithm-ideab4

Skipping attention mass ≤ δ of the total changes the output by at most 2δM

For all $\htmlData{mx=1}{m, k \in \htmlData{mx=0}{\mathbb{N}}}$ , $\htmlData{mx=5}{a, x : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{m}} \mathrel{\htmlData{mx=4}{\to}} \htmlData{mx=3}{\mathbb{R}}}$ , $\htmlData{mx=5}{b, y : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{k}} \mathrel{\htmlData{mx=4}{\to}} \htmlData{mx=3}{\mathbb{R}}}$ and $\htmlData{mx=1}{\delta, M \in \htmlData{mx=0}{\mathbb{R}}}$ , if $0 \mathrel{\htmlData{mx=1}{\leq}} \htmlData{mx=0}{M}$ , $\htmlData{mx=4}{\forall i,\ 0 \mathrel{\htmlData{mx=3}{\leq}} \htmlData{mx=2}{\htmlData{mx=0}{a}(\htmlData{mx=1}{i})}}$ , $\htmlData{mx=4}{\forall j,\ 0 \mathrel{\htmlData{mx=3}{\leq}} \htmlData{mx=2}{\htmlData{mx=0}{b}(\htmlData{mx=1}{j})}}$ , $0 \mathrel{\htmlData{mx=4}{<}} \htmlData{mx=3}{\sum_{i} \htmlData{mx=2}{\htmlData{mx=0}{a}(\htmlData{mx=1}{i})}}$ , $\htmlData{mx=3}{\sum_{j} \htmlData{mx=2}{\htmlData{mx=0}{b}(\htmlData{mx=1}{j})}} \mathrel{\htmlData{mx=15}{\leq}} \htmlData{mx=14}{\htmlData{mx=4}{\delta} \cdot \left(\htmlData{mx=13}{\htmlData{mx=8}{\sum_{i} \htmlData{mx=7}{\htmlData{mx=5}{a}(\htmlData{mx=6}{i})}} + \htmlData{mx=12}{\sum_{j} \htmlData{mx=11}{\htmlData{mx=9}{b}(\htmlData{mx=10}{j})}}}\right)}$ , $\htmlData{mx=6}{\forall i,\ \htmlData{mx=3}{\left\lvert \htmlData{mx=2}{\htmlData{mx=0}{x}(\htmlData{mx=1}{i})} \right\rvert} \mathrel{\htmlData{mx=5}{\leq}} \htmlData{mx=4}{M}}$ and $\htmlData{mx=6}{\forall j,\ \htmlData{mx=3}{\left\lvert \htmlData{mx=2}{\htmlData{mx=0}{y}(\htmlData{mx=1}{j})} \right\rvert} \mathrel{\htmlData{mx=5}{\leq}} \htmlData{mx=4}{M}}$ , then

\begin{aligned} & \htmlData{mx=42}{\left\lvert \htmlData{mx=41}{\htmlData{mx=27}{\frac{\htmlData{mx=17}{\htmlData{mx=8}{\sum_{i} \htmlData{mx=7}{\htmlData{mx=3}{\htmlData{mx=1}{a}(\htmlData{mx=2}{i})} \cdot \htmlData{mx=6}{\htmlData{mx=4}{x}(\htmlData{mx=5}{i})}}} + \htmlData{mx=16}{\sum_{j} \htmlData{mx=15}{\htmlData{mx=11}{\htmlData{mx=9}{b}(\htmlData{mx=10}{j})} \cdot \htmlData{mx=14}{\htmlData{mx=12}{y}(\htmlData{mx=13}{j})}}}}}{\htmlData{mx=26}{\htmlData{mx=21}{\sum_{i} \htmlData{mx=20}{\htmlData{mx=18}{a}(\htmlData{mx=19}{i})}} + \htmlData{mx=25}{\sum_{j} \htmlData{mx=24}{\htmlData{mx=22}{b}(\htmlData{mx=23}{j})}}}}} - \htmlData{mx=40}{\frac{\htmlData{mx=35}{\sum_{i} \htmlData{mx=34}{\htmlData{mx=30}{\htmlData{mx=28}{a}(\htmlData{mx=29}{i})} \cdot \htmlData{mx=33}{\htmlData{mx=31}{x}(\htmlData{mx=32}{i})}}}}{\htmlData{mx=39}{\sum_{i} \htmlData{mx=38}{\htmlData{mx=36}{a}(\htmlData{mx=37}{i})}}}}} \right\rvert} \\ & \quad \htmlData{mx=0}{\leq} \htmlData{mx=46}{\htmlData{mx=44}{2\htmlData{mx=43}{\delta}} \cdot \htmlData{mx=45}{M}} \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: m, k range over natural numbers: the statement must hold for every choice. ((m k : 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 m. Lean writes f x; papers write f(x). (Fin m)
⟹Implies: If Fin(m), then ℝ. (Between types, A → B is the type of functions from A to B.) (Fin m → ℝ)
x : TVariables: a, x range over Fin m → ℝ: the statement must hold for every choice. ((a x : Fin m → ℝ))
≤At most: 0 is less than or equal to M. (0 ≤ M)
∀For all: The statement after the comma holds for every i. (∀ i, 0 ≤ a i)
<Less than: 0 is strictly smaller than ∑ i, a(i). (0 < ∑ i, a i)
a/bDivision: (∑ i, a(i) · x(i)) + (∑ j, b(j) · y(j)) divided by (∑ i, a(i)) + (∑ j, b(j)). In Lean, dividing by 0 gives 0. ((∑ i, a i * x i + ∑ j, b j * y j) / (∑ …)
-Subtraction: ((∑ i, a(i) · x(i)) + (∑ j, b(j) · y(j)))/((∑ i, a(i)) + (… minus (∑ i, a(i) · x(i))/(∑ i, a(i)). On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. ((∑ i, a i * x i + ∑ j, b j * y j) / (∑ …)
∑Summation: Add up a(i) for every i. (∑ i, a i)
|x|Absolute value: The size of x(i) without its sign. (|x i|)
RReal numbers: The real numbers: every point on the number line. (ℝ)
·Multiplication: δ times (∑ i, a(i)) + (∑ j, b(j)). (δ * (∑ i, a i + ∑ j, b j))
+Addition: ∑ i, a(i) plus ∑ j, b(j). (∑ i, a i + ∑ j, b j)

theorem tile-skip-output-error {m k : ℕ} (a x : Fin m → ℝ) (b y : Fin k → ℝ) (δ M : ℝ) (hM : 0 ≤ M) (ha : ∀ i, 0 ≤ a i) (hb : ∀ j, 0 ≤ b j) (hA : 0 < ∑ i, a i) (hδ : ∑ j, b j ≤ δ * (∑ i, a i + ∑ j, b j)) (hx : ∀ i, |x i| ≤ M) (hy : ∀ j, |y j| ≤ M) :
    |(∑ i, a i * x i + ∑ j, b j * y j) / (∑ i, a i + ∑ j, b j) - (∑ i, a i * x i) / ∑ i, a i| ≤ 2 * δ * M

Idea B4. Kept keys have nonnegative weights aᵢ (e^{score}) and values xᵢ; skipped keys have weights bⱼ and values yⱼ, all |values| ≤ M. If the skipped mass is at most δ of the total, the attention output computed from the kept keys alone is within 2δM of the full output. With δ below half a BF16 ulp the skip is invisible at BF16 resolution up to the factor 2M.

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: 38 lines of Lean

Skipping attention mass ≤ δ of the total changes the output by at most 2δM

How it was proved

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