Provedllmlarge-language-modelsformal-proofattention

Weight0 in raw exponential form

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=12}{=}} \htmlData{mx=11}{\frac{\htmlData{mx=5}{e^{\htmlData{mx=4}{s_{0}}}}}{\htmlData{mx=10}{\htmlData{mx=7}{e^{\htmlData{mx=6}{s_{0}}}} + \htmlData{mx=9}{e^{\htmlData{mx=8}{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))
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 exp(s0)/(exp(s0) + exp(s1)) are the same. (TransformerAttention.weight0 s0 s1 = Re…)
eˣExponential function: e raised to the power s0, where e ≈ 2.718. (Real.exp s0)
a/bDivision: exp(s0) divided by exp(s0) + exp(s1). In Lean, dividing by 0 gives 0. (Real.exp s0 / (Real.exp s0 + Real.exp s…)
+Addition: exp(s0) plus exp(s1). (Real.exp s0 + Real.exp s1)

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

weight0 s0 s1 = exp s0 / (exp s0 + exp s1): the definitional unfolding.

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: 13 lines of Lean