Provedllmlarge-language-modelsformal-proofoptimization

The floor parameter enters additively

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

\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}{x})} \mathrel{\htmlData{mx=16}{=}} \htmlData{mx=15}{\htmlData{mx=13}{\htmlData{mx=8}{\operatorname{qloss}}(\htmlData{mx=9}{a}, \htmlData{mx=10}{m}, \htmlData{mx=11}{c}, \htmlData{mx=12}{x})} + \htmlData{mx=14}{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', x range over real numbers: the statement must hold for every choice. ((a m c c' x : 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', x. Lean writes f x; papers write f(x). (LLMOptim.qloss a m (c + c') x)
=Equality: qloss(a, m, c + c', x) and qloss(a, m, c, x) + c' are the same. (LLMOptim.qloss a m (c + c') x = LLMOpti…)
+Addition: c plus c'. (c + c')

theorem qloss-additive-floor (a m c c' x : ℝ) :
    LLMOptim.qloss a m (c + c') x = LLMOptim.qloss a m c x + c'

qloss with floor c + c' equals qloss with floor c plus c'.

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