Provedllmlarge-language-modelsformal-proofoptimization

The loss is symmetric in equidistant points

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

\htmlData{mx=7}{\htmlData{mx=0}{\operatorname{qloss}}(\htmlData{mx=1}{a}, \htmlData{mx=2}{m}, \htmlData{mx=3}{c}, \htmlData{mx=6}{\htmlData{mx=4}{m} + \htmlData{mx=5}{d}})} \mathrel{\htmlData{mx=16}{=}} \htmlData{mx=15}{\htmlData{mx=8}{\operatorname{qloss}}(\htmlData{mx=9}{a}, \htmlData{mx=10}{m}, \htmlData{mx=11}{c}, \htmlData{mx=14}{\htmlData{mx=12}{m} - \htmlData{mx=13}{d}})}

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, d range over real numbers: the statement must hold for every choice. ((a m c d : 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, m + d. Lean writes f x; papers write f(x). (LLMOptim.qloss a m c (m + d))
=Equality: qloss(a, m, c, m + d) and qloss(a, m, c, m − d) are the same. (LLMOptim.qloss a m c (m + d) = LLMOptim…)
+Addition: m plus d. (m + d)
-Subtraction: m minus d. On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. (m - d)

theorem opt-qloss-sym-diff (a m c d : ℝ) :
    LLMOptim.qloss a m c (m + d) = LLMOptim.qloss a m c (m - d)

qloss a m c (m + d) = qloss a m c (m − d).

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