proofs
Provedunbiased-attentionrandomized-algorithmssoftmaxhardness

OV gadget NO case: all ⟨a,b⟩ ≥ 1 and C ≥ 2 log(4n) force attention output ≤ 1/4

For all n∈N\htmlData{mx=1}{n \in \htmlData{mx=0}{\mathbb{N}}}, ip:Fin⁡ n→N\htmlData{mx=5}{\mathit{ip} : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{n}} \mathrel{\htmlData{mx=4}{\to}} \htmlData{mx=3}{\mathbb{N}}} and C∈R\htmlData{mx=1}{C \in \htmlData{mx=0}{\mathbb{R}}}, if 0<n0 \mathrel{\htmlData{mx=1}{<}} \htmlData{mx=0}{n}, 2log⁡(4n)≤C\htmlData{mx=3}{2\htmlData{mx=2}{\log (\htmlData{mx=1}{4\htmlData{mx=0}{n}})}} \mathrel{\htmlData{mx=5}{\leq}} \htmlData{mx=4}{C} and ∀k, 1≤ip(k)\htmlData{mx=4}{\forall k,\ 1 \mathrel{\htmlData{mx=3}{\leq}} \htmlData{mx=2}{\htmlData{mx=0}{\mathit{ip}}(\htmlData{mx=1}{k})}}, then

∑ke−C⋅ip(k)e−C2+∑ke−C⋅ip(k)≤14\begin{aligned} & \htmlData{mx=22}{\frac{\htmlData{mx=8}{\sum_{k} \htmlData{mx=7}{e^{\htmlData{mx=6}{\htmlData{mx=2}{-\htmlData{mx=1}{C}} \cdot \htmlData{mx=5}{\htmlData{mx=3}{\mathit{ip}}(\htmlData{mx=4}{k})}}}}}}{\htmlData{mx=21}{\htmlData{mx=12}{e^{\htmlData{mx=11}{\frac{\htmlData{mx=10}{-\htmlData{mx=9}{C}}}{2}}}} + \htmlData{mx=20}{\sum_{k} \htmlData{mx=19}{e^{\htmlData{mx=18}{\htmlData{mx=14}{-\htmlData{mx=13}{C}} \cdot \htmlData{mx=17}{\htmlData{mx=15}{\mathit{ip}}(\htmlData{mx=16}{k})}}}}}}}} \\ & \quad \htmlData{mx=0}{\leq} \htmlData{mx=23}{\frac{1}{4}} \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 ℕ. (Between types, A → B is the type of functions from A to B.) (Fin n → ℕ)
RReal numbers
The real numbers: every point on the number line. (ℝ)
<Less than
0 is strictly smaller than n. (0 < n)
≤At most
2 · log(4 · n) is less than or equal to C. (2 * Real.log (4 * n) ≤ C)
∀For all
The statement after the comma holds for every k. (∀ k, 1 ≤ ip k)
∑Summation
Add up exp(−C · ip(k)) for every k. (∑ k, Real.exp (-C * ip k))
logNatural logarithm
The logarithm of 4 · n to base e: the power e must be raised to to give 4 · n. (Real.log (4 * n))
eˣExponential function
e raised to the power −C · ip(k), where e ≈ 2.718. (Real.exp (-C * ip k))
·Multiplication
4 times n. (4 * n)
−xNegation
The negative of C. (-C)
+Addition
exp(−C/2) plus ∑ k, exp(−C · ip(k)). (Real.exp (-C / 2) + ∑ k, Real.exp (-C *…)
a/bDivision
−C divided by 2. In Lean, dividing by 0 gives 0. (-C / 2)

If no orthogonal partner exists every key weight is ≤ e^{−C}, so S ≤ n e^{−C} and the output is ≤ n e^{−C/2} ≤ 1/4. Together with the YES case this is a constant gap on a single attention row. Part of 'Unbiased randomized attention: cost x variance thresholds' (open theory problem).

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
42 lines of Lean
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