Provedalphabaguettealgorithm-ideab4

Per-dimension key bounds give an upper bound on every q·k in a tile

For all $\htmlData{mx=1}{n \in \htmlData{mx=0}{\mathbb{N}}}$ and $\htmlData{mx=5}{q, k, \mathit{lo}, \mathit{hi} : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{n}} \mathrel{\htmlData{mx=4}{\to}} \htmlData{mx=3}{\mathbb{R}}}$ , if $\htmlData{mx=7}{\forall j,\ \htmlData{mx=2}{\htmlData{mx=0}{\mathit{lo}}(\htmlData{mx=1}{j})} \mathrel{\htmlData{mx=6}{\leq}} \htmlData{mx=5}{\htmlData{mx=3}{k}(\htmlData{mx=4}{j})}}$ and $\htmlData{mx=7}{\forall j,\ \htmlData{mx=2}{\htmlData{mx=0}{k}(\htmlData{mx=1}{j})} \mathrel{\htmlData{mx=6}{\leq}} \htmlData{mx=5}{\htmlData{mx=3}{\mathit{hi}}(\htmlData{mx=4}{j})}}$ , then

\htmlData{mx=7}{\sum_{j} \htmlData{mx=6}{\htmlData{mx=2}{\htmlData{mx=0}{q}(\htmlData{mx=1}{j})} \cdot \htmlData{mx=5}{\htmlData{mx=3}{k}(\htmlData{mx=4}{j})}}} \mathrel{\htmlData{mx=24}{\leq}} \htmlData{mx=23}{\sum_{j} \htmlData{mx=22}{\max(\htmlData{mx=14}{\htmlData{mx=10}{\htmlData{mx=8}{q}(\htmlData{mx=9}{j})} \cdot \htmlData{mx=13}{\htmlData{mx=11}{\mathit{lo}}(\htmlData{mx=12}{j})}}, \htmlData{mx=21}{\htmlData{mx=17}{\htmlData{mx=15}{q}(\htmlData{mx=16}{j})} \cdot \htmlData{mx=20}{\htmlData{mx=18}{\mathit{hi}}(\htmlData{mx=19}{j})}})}}

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: q, k, lo, hi range over Fin n → ℝ: the statement must hold for every choice. ((q k lo hi : Fin n → ℝ))
∀For all: The statement after the comma holds for every j. (∀ j, lo j ≤ k j)
∑Summation: Add up q(j) · k(j) for every j. (∑ j, q j * k j)
≤At most: lo(j) is less than or equal to k(j). (lo j ≤ k j)
·Multiplication: q(j) times k(j). (q j * k j)
RReal numbers: The real numbers: every point on the number line. (ℝ)
maxMaximum: The larger of q(j) · lo(j) and q(j) · hi(j). (max (q j * lo j) (q j * hi j))

theorem tile-box-score-bound {n : ℕ} (q k lo hi : Fin n → ℝ) (hlo : ∀ j, lo j ≤ k j) (hhi : ∀ j, k j ≤ hi j) :
    ∑ j, q j * k j ≤ ∑ j, max (q j * lo j) (q j * hi j)

Idea B4 (certified attention-tile skipping). If each key in a tile satisfies lo ≤ k ≤ hi coordinatewise, then q·k ≤ Σⱼ max(qⱼ·loⱼ, qⱼ·hiⱼ). The bound costs 2·64 multiplies per tile instead of 64 per key.

How it was proved

Proved: 3 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: 15 lines of Lean

Per-dimension key bounds give an upper bound on every q·k in a tile

How it was proved

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