Provedbalanced-conversionconversion-theoremattention-compute-lawlower-bound

Balanced split or heavy key: if every key has mass ≤ 3/4, some prefix has mass in [1/8, 7/8]

For all $\htmlData{mx=1}{n \in \htmlData{mx=0}{\mathbb{N}}}$ and $\htmlData{mx=3}{m : \htmlData{mx=0}{\mathbb{N}} \mathrel{\htmlData{mx=2}{\to}} \htmlData{mx=1}{\mathbb{R}}}$ , if $\htmlData{mx=4}{\forall j,\ 0 \mathrel{\htmlData{mx=3}{\leq}} \htmlData{mx=2}{\htmlData{mx=0}{m}(\htmlData{mx=1}{j})}}$ , $\htmlData{mx=4}{\sum_{j=0}^{\htmlData{mx=0}{n} - 1} \htmlData{mx=3}{\htmlData{mx=1}{m}(\htmlData{mx=2}{j})}} \mathrel{\htmlData{mx=5}{=}} 1$ and $\htmlData{mx=9}{\forall j,\ \htmlData{mx=0}{j} \mathrel{\htmlData{mx=2}{<}} \htmlData{mx=1}{n} \mathrel{\htmlData{mx=8}{\implies}} \htmlData{mx=5}{\htmlData{mx=3}{m}(\htmlData{mx=4}{j})} \mathrel{\htmlData{mx=7}{\leq}} \htmlData{mx=6}{\frac{3}{4}}}$ , then

\htmlData{mx=19}{\exists i,\ \htmlData{mx=0}{i} \mathrel{\htmlData{mx=2}{\leq}} \htmlData{mx=1}{n} \mathbin{\htmlData{mx=18}{\land}} \htmlData{mx=3}{\frac{1}{8}} \mathrel{\htmlData{mx=9}{\leq}} \htmlData{mx=8}{\sum_{j=0}^{\htmlData{mx=4}{i} - 1} \htmlData{mx=7}{\htmlData{mx=5}{m}(\htmlData{mx=6}{j})}} \mathbin{\htmlData{mx=17}{\land}} \htmlData{mx=14}{\sum_{j=0}^{\htmlData{mx=10}{i} - 1} \htmlData{mx=13}{\htmlData{mx=11}{m}(\htmlData{mx=12}{j})}} \mathrel{\htmlData{mx=16}{\leq}} \htmlData{mx=15}{\frac{7}{8}}}

What the notation means · or point at any symbol

NNatural numbers: The natural numbers 0, 1, 2, …. In Lean they start at 0. (ℕ)
⟹Implies: If ℕ, then ℝ. (Between types, A → B is the type of functions from A to B.) (ℕ → ℝ)
∀For all: The statement after the comma holds for every j. (∀ j, 0 ≤ m j)
∑Summation: Add up m(j) for j = 0, 1, …, n − 1. (∑ j ∈ Finset.range n, m j)
=Equality: ∑ j, m(j) and 1 are the same. (∑ j ∈ Finset.range n, m j = 1)
∃There exists: At least one i makes the statement after the comma true. (∃ i, i ≤ n ∧ 1 / 8 ≤ ∑ j ∈ Finset.range…)
RReal numbers: The real numbers: every point on the number line. (ℝ)
≤At most: 0 is less than or equal to m(j). (0 ≤ m j)
<Less than: j is strictly smaller than n. (j < n)
f(x)Function application: m applied to j. Lean writes f x; papers write f(x). (m j)
∧And: Both hold: 1/8 ≤ (∑ j, m(j)), and (∑ j, m(j)) ≤ 7/8. (1 / 8 ≤ ∑ j ∈ Finset.range i, m j ∧ ∑ j…)
a/bDivision: 3 divided by 4. In Lean, dividing by 0 gives 0. (3 / 4)

theorem bc2-balanced-prefix-exists (n : ℕ) (m : ℕ → ℝ) (hm : ∀ j, 0 ≤ m j) (hsum : ∑ j ∈ Finset.range n, m j = 1) (hmax : ∀ j, j < n → m j ≤ 3 / 4) :
    ∃ i, i ≤ n ∧ 1 / 8 ≤ ∑ j ∈ Finset.range i, m j ∧ ∑ j ∈ Finset.range i, m j ≤ 7 / 8

The mass-balanced descent always finds a balanced prefix split by binary search on prefix masses unless a single key is heavy (mass > 3/4), in which case that key exits as a leaf. Part of Balanced Conversion: the Conversion Theorem's two-way descent factor is exactly 1 + (p−q)²/(q(1−q)), so additive oracle error is second order at mass-balanced nodes; a mass-balanced descent converts any deterministic additive-ε oracle with loss n^{O(ε²)}, and the biased product family shows this is tight. Quantifies the last residual class of the Attention Compute Law.

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: 28 lines of Lean

Balanced split or heavy key: if every key has mass ≤ 3/4, some prefix has mass in [1/8, 7/8]

How it was proved

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