Provedunbiased-attentionrandomized-algorithmssoftmaxhardness

Tensor powers of 0/1 vectors are 0/1 vectors

For all $\htmlData{mx=1}{\iota \in \htmlData{mx=0}{\operatorname{Type}}}$ , $\htmlData{mx=1}{k \in \htmlData{mx=0}{\mathbb{N}}}$ , $\htmlData{mx=3}{u : \htmlData{mx=0}{\iota} \mathrel{\htmlData{mx=2}{\to}} \htmlData{mx=1}{\mathbb{R}}}$ and $\htmlData{mx=5}{f : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{k}} \mathrel{\htmlData{mx=4}{\to}} \htmlData{mx=3}{\iota}}$ , if $\htmlData{mx=9}{\forall i,\ \htmlData{mx=2}{\htmlData{mx=0}{u}(\htmlData{mx=1}{i})} \mathrel{\htmlData{mx=3}{=}} 0 \mathbin{\htmlData{mx=8}{\lor}} \htmlData{mx=6}{\htmlData{mx=4}{u}(\htmlData{mx=5}{i})} \mathrel{\htmlData{mx=7}{=}} 1}$ , then

\left(\htmlData{mx=5}{\prod_{i} \htmlData{mx=4}{\htmlData{mx=0}{u}(\htmlData{mx=3}{\htmlData{mx=1}{f}(\htmlData{mx=2}{i})})}}\right) \mathrel{\htmlData{mx=6}{=}} 0 \mathbin{\htmlData{mx=14}{\lor}} \left(\htmlData{mx=12}{\prod_{i} \htmlData{mx=11}{\htmlData{mx=7}{u}(\htmlData{mx=10}{\htmlData{mx=8}{f}(\htmlData{mx=9}{i})})}}\right) \mathrel{\htmlData{mx=13}{=}} 1

What the notation means · or point at any symbol

Type: Type: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (Type)
NNatural numbers: The natural numbers 0, 1, 2, …. In Lean they start at 0. (ℕ)
⟹Implies: If ι, then ℝ. (Between types, A → B is the type of functions from A to B.) (ι → ℝ)
∀For all: The statement after the comma holds for every i. (∀ i, u i = 0 ∨ u i = 1)
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)
∏Product: Multiply together u(f(i)) for every i. (∏ i, u (f i))
RReal numbers: The real numbers: every point on the number line. (ℝ)
f(x)Function application: u applied to i. Lean writes f x; papers write f(x). (u i)
=Equality: u(i) and 0 are the same. (u i = 0)
∨Or: At least one holds: u(i) = 0, or u(i) = 1. (u i = 0 ∨ u i = 1)

theorem ua-tensor-power-boolean {ι : Type} (k : ℕ) (u : ι → ℝ) (hu : ∀ i, u i = 0 ∨ u i = 1) (f : Fin k → ι) :
    (∏ i, u (f i)) = 0 ∨ (∏ i, u (f i)) = 1

Each coordinate ∏ u(f i) of u^{⊗k} is 0 or 1, so the tensored Rubinstein instance stays a 0/1 instance (keys, queries and the Hamming/inner-product embeddings stay valid). Part of 'Unbiased randomized attention: cost x variance thresholds', round 10 (isolation grid and Boolean tensor powers).

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: 15 lines of Lean

Tensor powers of 0/1 vectors are 0/1 vectors

How it was proved

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