Provedllmlarge-language-modelsformal-proofattention

Tied-score attention is homogeneous in the value pair

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

\htmlData{mx=9}{\htmlData{mx=0}{\operatorname{attend2}}(\htmlData{mx=1}{s}, \htmlData{mx=2}{s}, \htmlData{mx=5}{\htmlData{mx=3}{c} \cdot \htmlData{mx=4}{v_{0}}}, \htmlData{mx=8}{\htmlData{mx=6}{c} \cdot \htmlData{mx=7}{v_{1}}})} \mathrel{\htmlData{mx=18}{=}} \htmlData{mx=17}{\htmlData{mx=10}{c} \cdot \htmlData{mx=16}{\htmlData{mx=11}{\operatorname{attend2}}(\htmlData{mx=12}{s}, \htmlData{mx=13}{s}, \htmlData{mx=14}{v_{0}}, \htmlData{mx=15}{v_{1}})}}

What the notation means · or point at any symbol

RReal numbers: The real numbers: every point on the number line. (ℝ)
x ∈ TVariables: s, c, v0, v1 range over real numbers: the statement must hold for every choice. ((s c v0 v1 : ℝ))
attend2: TransformerAttention.attend2: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (TransformerAttention.attend2)
f(x)Function application: TransformerAttention.attend2 applied to s, s, c · v0, c · v1. Lean writes f x; papers write f(x). (TransformerAttention.attend2 s s (c * v…)
=Equality: attend2(s, s, c · v0, c · v1) and c · attend2(s, s, v0, v1) are the same. (TransformerAttention.attend2 s s (c * v…)
·Multiplication: c times v0. (c * v0)

theorem attn2-attend2-tie-smul (s c v0 v1 : ℝ) :
    TransformerAttention.attend2 s s (c * v0) (c * v1) = c * TransformerAttention.attend2 s s v0 v1

attend2 s s (c·v0) (c·v1) = c · attend2 s s v0 v1.

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: 14 lines of Lean