Provedunbiased-attentionrandomized-algorithmssoftmaxhardness

Range-aware gap, YES: one key scoring ≥ baseline + g (g ≥ 0) forces output ≥ 1/2

For all $\htmlData{mx=1}{n \in \htmlData{mx=0}{\mathbb{N}}}$ , $\htmlData{mx=5}{s : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{n}} \mathrel{\htmlData{mx=4}{\to}} \htmlData{mx=3}{\mathbb{R}}}$ , $\htmlData{mx=1}{b, g \in \htmlData{mx=0}{\mathbb{R}}}$ and $\htmlData{mx=3}{j \in \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{n}}}$ , if $0 \mathrel{\htmlData{mx=1}{\leq}} \htmlData{mx=0}{g}$ and $\htmlData{mx=2}{\htmlData{mx=0}{b} + \htmlData{mx=1}{g}} \mathrel{\htmlData{mx=6}{\leq}} \htmlData{mx=5}{\htmlData{mx=3}{s}(\htmlData{mx=4}{j})}$ , then

\htmlData{mx=0}{\frac{1}{2}} \mathrel{\htmlData{mx=15}{\leq}} \htmlData{mx=14}{\frac{\htmlData{mx=5}{\sum_{k} \htmlData{mx=4}{e^{\htmlData{mx=3}{\htmlData{mx=1}{s}(\htmlData{mx=2}{k})}}}}}{\htmlData{mx=13}{\htmlData{mx=7}{e^{\htmlData{mx=6}{b}}} + \htmlData{mx=12}{\sum_{k} \htmlData{mx=11}{e^{\htmlData{mx=10}{\htmlData{mx=8}{s}(\htmlData{mx=9}{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)
x ∈ TVariables: b, g range over real numbers: the statement must hold for every choice. ((b g : ℝ))
≤At most: 0 is less than or equal to g. (0 ≤ g)
+Addition: b plus g. (b + g)
a/bDivision: 1 divided by 2. In Lean, dividing by 0 gives 0. (1 / 2)
RReal numbers: The real numbers: every point on the number line. (ℝ)
∑Summation: Add up exp(s(k)) for every k. (∑ k, Real.exp (s k))
eˣExponential function: e raised to the power s(k), where e ≈ 2.718. (Real.exp (s k))

theorem ua-gap-general-yes {n : ℕ} (s : Fin n → ℝ) (b g : ℝ) (hg : 0 ≤ g) (j : Fin n) (hj : b + g ≤ s j) :
    1 / 2 ≤ (∑ k, Real.exp (s k)) / (Real.exp b + ∑ k, Real.exp (s k))

Generalizes the OV gadget to arbitrary real scores: value-1 keys s_k, a value-0 baseline key at score b. If some s_j ≥ b + g with g ≥ 0, the output Σe^{s}/(e^b + Σe^{s}) is ≥ 1/2. Only the gap g matters, not the absolute range. Part of 'Unbiased randomized attention: cost x variance thresholds' (open theory problem), round 2.

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

Range-aware gap, YES: one key scoring ≥ baseline + g (g ≥ 0) forces output ≥ 1/2

How it was proved

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