Provedllmlarge-language-modelsformal-proofdecoding

Greedy two-logit choice returns 0 when the first does not lose

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

\htmlData{mx=3}{\lnot \htmlData{mx=0}{x} \mathrel{\htmlData{mx=2}{<}} \htmlData{mx=1}{y}} \mathrel{\htmlData{mx=9}{\implies}} \htmlData{mx=7}{\htmlData{mx=4}{\operatorname{argmax2}}(\htmlData{mx=5}{x}, \htmlData{mx=6}{y})} \mathrel{\htmlData{mx=8}{=}} 0

What the notation means · or point at any symbol

RReal numbers: The real numbers: every point on the number line. (ℝ)
x ∈ TVariables: x, y range over real numbers: the statement must hold for every choice. ((x y : ℝ))
¬Not: It is not the case that x < y. (¬ x < y)
⟹Implies: If ¬(x < y), then argmax2(x, y) = 0. (Between types, A → B is the type of functions from A to B.) (¬ x < y → LLMDecode.argmax2 x y = 0)
<Less than: x is strictly smaller than y. (x < y)
argmax2: LLMDecode.argmax2: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (LLMDecode.argmax2)
f(x)Function application: LLMDecode.argmax2 applied to x, y. Lean writes f x; papers write f(x). (LLMDecode.argmax2 x y)
=Equality: argmax2(x, y) and 0 are the same. (LLMDecode.argmax2 x y = 0)

theorem dec-argmax2-left (x y : ℝ) :
    ¬ x < y → LLMDecode.argmax2 x y = 0

If ¬(x < y) then argmax2 x y = 0: leftmost tie-breaking.

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