Provedllmlarge-language-modelsformal-proofattention

First weight is at least one half when it wins the tie-break

For all $\htmlData{mx=1}{s_{0}, s_{1} \in \htmlData{mx=0}{\mathbb{R}}}$ :

\htmlData{mx=0}{s_{1}} \mathrel{\htmlData{mx=2}{\leq}} \htmlData{mx=1}{s_{0}} \mathrel{\htmlData{mx=9}{\implies}} \htmlData{mx=3}{\frac{1}{2}} \mathrel{\htmlData{mx=8}{\leq}} \htmlData{mx=7}{\htmlData{mx=4}{\operatorname{weight0}}(\htmlData{mx=5}{s_{0}}, \htmlData{mx=6}{s_{1}})}

What the notation means · or point at any symbol

RReal numbers: The real numbers: every point on the number line. (ℝ)
x ∈ TVariables: s0, s1 range over real numbers: the statement must hold for every choice. ((s0 s1 : ℝ))
≤At most: s1 is less than or equal to s0. (s1 ≤ s0)
⟹Implies: If s1 ≤ s0, then 1/2 ≤ weight0(s0, s1). (Between types, A → B is the type of functions from A to B.) (s1 ≤ s0 → 1 / 2 ≤ TransformerAttention.…)
a/bDivision: 1 divided by 2. In Lean, dividing by 0 gives 0. (1 / 2)
weight0: TransformerAttention.weight0: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (TransformerAttention.weight0)
f(x)Function application: TransformerAttention.weight0 applied to s0, s1. Lean writes f x; papers write f(x). (TransformerAttention.weight0 s0 s1)

theorem attn2-weight0-ge-half (s0 s1 : ℝ) :
    s1 ≤ s0 → 1 / 2 ≤ TransformerAttention.weight0 s0 s1

If s1 ≤ s0 then weight0 ≥ 1/2.

How it was proved

Proved: 28 September 2026
By: Ajax, with the agent “ompmuse”
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