Provedllmlarge-language-modelsformal-proofoptimization

Loss at minimizer with split floor

For all $\htmlData{mx=1}{a, m, c, c' \in \htmlData{mx=0}{\mathbb{R}}}$ :

\htmlData{mx=7}{\htmlData{mx=0}{\operatorname{qloss}}(\htmlData{mx=1}{a}, \htmlData{mx=2}{m}, \htmlData{mx=5}{\htmlData{mx=3}{c} + \htmlData{mx=4}{c'}}, \htmlData{mx=6}{m})} \mathrel{\htmlData{mx=11}{=}} \htmlData{mx=10}{\htmlData{mx=8}{c} + \htmlData{mx=9}{c'}}

What the notation means · or point at any symbol

RReal numbers: The real numbers: every point on the number line. (ℝ)
x ∈ TVariables: a, m, c, c' range over real numbers: the statement must hold for every choice. ((a m c c' : ℝ))
qloss: LLMOptim.qloss: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (LLMOptim.qloss)
f(x)Function application: LLMOptim.qloss applied to a, m, c + c', m. Lean writes f x; papers write f(x). (LLMOptim.qloss a m (c + c') m)
=Equality: qloss(a, m, c + c', m) and c + c' are the same. (LLMOptim.qloss a m (c + c') m = c + c')
+Addition: c plus c'. (c + c')

theorem opt-qloss-at-min-plus (a m c c' : ℝ) :
    LLMOptim.qloss a m (c + c') m = c + c'

qloss a m (c + c') m = c + c'.

How it was proved

Proved: 27 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