Provedunbiased-attentionrandomized-algorithmssoftmaxgradient-hardness

Far-key mass: all x ≤ s < 1 and β = log(8n)/(1−s) ⇒ Σ e^{β x_k} ≤ e^β/8

For all $\htmlData{mx=1}{n \in \htmlData{mx=0}{\mathbb{N}}}$ , $\htmlData{mx=5}{x : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{n}} \mathrel{\htmlData{mx=4}{\to}} \htmlData{mx=3}{\mathbb{R}}}$ and $\htmlData{mx=1}{s \in \htmlData{mx=0}{\mathbb{R}}}$ , if $0 \mathrel{\htmlData{mx=1}{<}} \htmlData{mx=0}{n}$ , $\htmlData{mx=0}{s} \mathrel{\htmlData{mx=1}{<}} 1$ , $0 \mathrel{\htmlData{mx=1}{\leq}} \htmlData{mx=0}{s}$ and $\htmlData{mx=5}{\forall k,\ \htmlData{mx=2}{\htmlData{mx=0}{x}(\htmlData{mx=1}{k})} \mathrel{\htmlData{mx=4}{\leq}} \htmlData{mx=3}{s}}$ , then

\begin{aligned} & \htmlData{mx=12}{\sum_{k} \htmlData{mx=11}{e^{\htmlData{mx=10}{\htmlData{mx=6}{\frac{\htmlData{mx=3}{\log (\htmlData{mx=2}{8\htmlData{mx=1}{n}})}}{\htmlData{mx=5}{1 - \htmlData{mx=4}{s}}}} \cdot \htmlData{mx=9}{\htmlData{mx=7}{x}(\htmlData{mx=8}{k})}}}}} \\ & \quad \htmlData{mx=0}{\leq} \htmlData{mx=20}{\frac{\htmlData{mx=19}{e^{\htmlData{mx=18}{\frac{\htmlData{mx=15}{\log (\htmlData{mx=14}{8\htmlData{mx=13}{n}})}}{\htmlData{mx=17}{1 - \htmlData{mx=16}{s}}}}}}}{8}} \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)
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 n. Lean writes f x; papers write f(x). (Fin n)
⟹Implies: If Fin(n), then ℝ. (Between types, A → B is the type of functions from A to B.) (Fin n → ℝ)
<Less than: 0 is strictly smaller than n. (0 < n)
≤At most: 0 is less than or equal to s. (0 ≤ s)
∀For all: The statement after the comma holds for every k. (∀ k, x k ≤ s)
∑Summation: Add up exp(log(8 · n)/(1 − s) · x(k)) for every k. (∑ k, Real.exp (Real.log (8 * n) / (1 - …)
eˣExponential function: e raised to the power log(8 · n)/(1 − s) · x(k), where e ≈ 2.718. (Real.exp (Real.log (8 * n) / (1 - s) * …)
RReal numbers: The real numbers: every point on the number line. (ℝ)
logNatural logarithm: The logarithm of 8 · n to base e: the power e must be raised to to give 8 · n. (Real.log (8 * n))
a/bDivision: log(8 · n) divided by 1 − s. In Lean, dividing by 0 gives 0. (Real.log (8 * n) / (1 - s))
·Multiplication: 8 times n. (8 * n)
-Subtraction: 1 minus s. On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. (1 - s)

theorem ua-far-mass-bound {n : ℕ} (x : Fin n → ℝ) (s : ℝ) (hn : 0 < n) (hs : s < 1) (hs0 : 0 ≤ s) (hx : ∀ k, x k ≤ s) :
    ∑ k, Real.exp (Real.log (8 * n) / (1 - s) * x k) ≤ Real.exp (Real.log (8 * n) / (1 - s)) / 8

Gives both the NO-row bound (o ≤ 1/8) and the isolated YES row's upper bound S ≤ (9/8)e^β. Score range stays ≤ β = ln(8n)/(1−s) → α ln n after tensoring. Part of 'Unbiased randomized attention: cost x variance thresholds', round 7 (gradient hardness via isolation).

How it was proved

Proved: 4 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: 29 lines of Lean

Far-key mass: all x ≤ s < 1 and β = log(8n)/(1−s) ⇒ Σ e^{β x_k} ≤ e^β/8

How it was proved

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