Provedllmlarge-language-modelsformal-proofattention

First weight as one over one plus the negative logit gap

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

\htmlData{mx=3}{\htmlData{mx=0}{\operatorname{weight0}}(\htmlData{mx=1}{s_{0}}, \htmlData{mx=2}{s_{1}})} \mathrel{\htmlData{mx=10}{=}} \htmlData{mx=9}{\frac{1}{\htmlData{mx=8}{1 + \htmlData{mx=7}{e^{\htmlData{mx=6}{\htmlData{mx=4}{s_{1}} - \htmlData{mx=5}{s_{0}}}}}}}}

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))
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)
=Equality: weight0(s0, s1) and 1/(1 + exp(s1 − s0)) are the same. (TransformerAttention.weight0 s0 s1 = 1 …)
a/bDivision: 1 divided by 1 + exp(s1 − s0). In Lean, dividing by 0 gives 0. (1 / (1 + Real.exp (s1 - s0)))
+Addition: 1 plus exp(s1 − s0). (1 + Real.exp (s1 - s0))
eˣExponential function: e raised to the power s1 − s0, where e ≈ 2.718. (Real.exp (s1 - s0))
-Subtraction: s1 minus s0. On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. (s1 - s0)

theorem attn2-weight0-inv-form (s0 s1 : ℝ) :
    TransformerAttention.weight0 s0 s1 = 1 / (1 + Real.exp (s1 - s0))

weight0 s0 s1 = 1 / (1 + exp (s1 - s0)): softmax depends only on the score difference.

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