Provedunbiased-attentionrandomized-algorithmssoftmaxhardness

Sharp gadget YES: baseline key AT the top score β; a key with x = 1 forces output ≥ 1/2

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

\htmlData{mx=0}{\frac{1}{2}} \mathrel{\htmlData{mx=19}{\leq}} \htmlData{mx=18}{\frac{\htmlData{mx=7}{\sum_{k} \htmlData{mx=6}{e^{\htmlData{mx=5}{\htmlData{mx=1}{\beta} \cdot \htmlData{mx=4}{\htmlData{mx=2}{x}(\htmlData{mx=3}{k})}}}}}}{\htmlData{mx=17}{\htmlData{mx=9}{e^{\htmlData{mx=8}{\beta}}} + \htmlData{mx=16}{\sum_{k} \htmlData{mx=15}{e^{\htmlData{mx=14}{\htmlData{mx=10}{\beta} \cdot \htmlData{mx=13}{\htmlData{mx=11}{x}(\htmlData{mx=12}{k})}}}}}}}}

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)
=Equality: x(j) and 1 are the same. (x j = 1)
a/bDivision: 1 divided by 2. In Lean, dividing by 0 gives 0. (1 / 2)
≤At most: 1/2 is less than or equal to (∑ k, exp(β · x(k)))/(exp(β) + (∑ k, exp(β · x(k)))). (1 / 2 ≤ (∑ k, Real.exp (β * x k)) / (Re…)
RReal numbers: The real numbers: every point on the number line. (ℝ)
∑Summation: Add up exp(β · x(k)) for every k. (∑ k, Real.exp (β * x k))
eˣExponential function: e raised to the power β · x(k), where e ≈ 2.718. (Real.exp (β * x k))
·Multiplication: β times x(k). (β * x k)
+Addition: exp(β) plus ∑ k, exp(β · x(k)). (Real.exp β + ∑ k, Real.exp (β * x k))

theorem ua-sharp-gap-yes {n : ℕ} (x : Fin n → ℝ) (β : ℝ) (j : Fin n) (hj : x j = 1) :
    1 / 2 ≤ (∑ k, Real.exp (β * x k)) / (Real.exp β + ∑ k, Real.exp (β * x k))

Scores β·x_k for value-1 keys, value-0 baseline key at score β. If some x_j = 1 the output is ≥ 1/2, for every β: the YES side costs no score range at all. Part of 'Unbiased randomized attention: cost x variance thresholds', round 3 (sharp threshold at score range ln n).

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: 16 lines of Lean

Sharp gadget YES: baseline key AT the top score β; a key with x = 1 forces output ≥ 1/2

How it was proved

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