Provedretrieval-speed-dualityhead-triageattentionduality

Retrieval needs range: a key with weight ≥ c forces score range ≥ log(c·n)

For all $\htmlData{mx=1}{n \in \htmlData{mx=0}{\mathbb{N}}}$ , $\htmlData{mx=5}{s : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{n}} \mathrel{\htmlData{mx=4}{\to}} \htmlData{mx=3}{\mathbb{R}}}$ , $\htmlData{mx=1}{R, c \in \htmlData{mx=0}{\mathbb{R}}}$ and $\htmlData{mx=3}{j \in \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{n}}}$ , if $0 \mathrel{\htmlData{mx=1}{<}} \htmlData{mx=0}{c}$ , $\htmlData{mx=9}{\forall i, j,\ \htmlData{mx=6}{\htmlData{mx=2}{\htmlData{mx=0}{s}(\htmlData{mx=1}{i})} - \htmlData{mx=5}{\htmlData{mx=3}{s}(\htmlData{mx=4}{j})}} \mathrel{\htmlData{mx=8}{\leq}} \htmlData{mx=7}{R}}$ and $\htmlData{mx=0}{c} \mathrel{\htmlData{mx=11}{\leq}} \htmlData{mx=10}{\frac{\htmlData{mx=4}{e^{\htmlData{mx=3}{\htmlData{mx=1}{s}(\htmlData{mx=2}{j})}}}}{\htmlData{mx=9}{\sum_{k} \htmlData{mx=8}{e^{\htmlData{mx=7}{\htmlData{mx=5}{s}(\htmlData{mx=6}{k})}}}}}}$ , then

\htmlData{mx=3}{\log (\htmlData{mx=2}{\htmlData{mx=0}{c} \cdot \htmlData{mx=1}{n}})} \mathrel{\htmlData{mx=5}{\leq}} \htmlData{mx=4}{R}

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: R, c range over real numbers: the statement must hold for every choice. ((R c : ℝ))
<Less than: 0 is strictly smaller than c. (0 < c)
∀For all: The statement after the comma holds for every i, j. (∀ i j, s i - s j ≤ R)
logNatural logarithm: The logarithm of c · n to base e: the power e must be raised to to give c · n. (Real.log (c * n))
eˣExponential function: e raised to the power s(j), where e ≈ 2.718. (Real.exp (s j))
a/bDivision: exp(s(j)) divided by ∑ k, exp(s(k)). In Lean, dividing by 0 gives 0. (Real.exp (s j) / ∑ k, Real.exp (s k))
-Subtraction: s(i) minus s(j). On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. (s i - s j)
≤At most: s(i) − s(j) is less than or equal to R. (s i - s j ≤ R)
RReal numbers: The real numbers: every point on the number line. (ℝ)
·Multiplication: c times n. (c * n)
∑Summation: Add up exp(s(k)) for every k. (∑ k, Real.exp (s k))

theorem rsd-retrieval-needs-range {n : ℕ} (s : Fin n → ℝ) (R c : ℝ) (hc : 0 < c) (hR : ∀ i j, s i - s j ≤ R) (j : Fin n) (hj : c ≤ Real.exp (s j) / ∑ k, Real.exp (s k)) :
    Real.log (c * n) ≤ R

Contrapositive of dilution. Combined with the ua-* threshold (range > ln n is SETH-hard to accelerate; range < ln n is cheap by sampling), per-head retrievability and accelerability are mutually exclusive up to an O(1) window around ln n. Part of 'Retrieval–Speed Duality: the score range that lets a head retrieve is the range that makes it hard to accelerate'.

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: 25 lines of Lean

Retrieval needs range: a key with weight ≥ c forces score range ≥ log(c·n)

How it was proved

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