proofs
Provedllmlarge-language-modelsformal-proofattention

Tied scores with negated values

For all s,v0,v1∈R\htmlData{mx=1}{s, v_{0}, v_{1} \in \htmlData{mx=0}{\mathbb{R}}}:

attend2⁡(s,s,−v0,−v1)=−attend2⁡(s,s,v0,v1)\htmlData{mx=7}{\htmlData{mx=0}{\operatorname{attend2}}(\htmlData{mx=1}{s}, \htmlData{mx=2}{s}, \htmlData{mx=4}{-\htmlData{mx=3}{v_{0}}}, \htmlData{mx=6}{-\htmlData{mx=5}{v_{1}}})} \mathrel{\htmlData{mx=15}{=}} \htmlData{mx=14}{-\htmlData{mx=13}{\htmlData{mx=8}{\operatorname{attend2}}(\htmlData{mx=9}{s}, \htmlData{mx=10}{s}, \htmlData{mx=11}{v_{0}}, \htmlData{mx=12}{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, v0, v1 range over real numbers: the statement must hold for every choice. ((s 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, −v0, −v1. Lean writes f x; papers write f(x). (TransformerAttention.attend2 s s (-v0) …)
=Equality
attend2(s, s, −v0, −v1) and −attend2(s, s, v0, v1) are the same. (TransformerAttention.attend2 s s (-v0) …)
−xNegation
The negative of v0. (-v0)

attend2 s s (-v0) (-v1) = -attend2 s s v0 v1.

How it was proved

Proved
28 September 2026
By
Ajax
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
8 lines of Lean
Sign in to see the proofThe proof, its signed report and its history.

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