Provedalphabaguettebaguette-zschedule

If each update is schedule-independent, the whole training trajectory is

For all $\htmlData{mx=1}{\Theta, B, \mathit{Sch} \in \htmlData{mx=0}{\operatorname{Type}}}$ , $\htmlData{mx=7}{U : \htmlData{mx=0}{\Theta} \mathrel{\htmlData{mx=6}{\to}} \htmlData{mx=1}{B} \mathrel{\htmlData{mx=5}{\to}} \htmlData{mx=2}{\mathit{Sch}} \mathrel{\htmlData{mx=4}{\to}} \htmlData{mx=3}{\Theta}}$ , $\htmlData{mx=1}{\theta_{0} \in \htmlData{mx=0}{\Theta}}$ , $\htmlData{mx=3}{\mathit{batch} : \htmlData{mx=0}{\mathbb{N}} \mathrel{\htmlData{mx=2}{\to}} \htmlData{mx=1}{B}}$ and $\htmlData{mx=3}{\sigma, \sigma' : \htmlData{mx=0}{\mathbb{N}} \mathrel{\htmlData{mx=2}{\to}} \htmlData{mx=1}{\mathit{Sch}}}$ , if $\htmlData{mx=11}{\forall \theta, b, s, s',\ \htmlData{mx=4}{\htmlData{mx=0}{U}(\htmlData{mx=1}{\theta}, \htmlData{mx=2}{b}, \htmlData{mx=3}{s})} \mathrel{\htmlData{mx=10}{=}} \htmlData{mx=9}{\htmlData{mx=5}{U}(\htmlData{mx=6}{\theta}, \htmlData{mx=7}{b}, \htmlData{mx=8}{s'})}}$ , then

\htmlData{mx=5}{\htmlData{mx=4}{\forall T \in \htmlData{mx=3}{\mathbb{N}},\ \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{rec}}(\htmlData{mx=1}{\operatorname{motive}})}}(\mathtt{:=\ fun\ \_\ =>\ {\Theta})\ {\theta}0\ (fun\ t\ {\theta}\ =>\ U\ {\theta}\ (batch\ t)\ ({\sigma}\ t))\ T\ =\ Nat.rec\ (motive\ :=\ fun\ \_\ =>\ {\Theta})\ {\theta}0\ (fun\ t\ {\theta}\ =>\ U\ {\theta}\ (batch\ t)\ ({\sigma}'\ t))\ T})}

What the notation means · or point at any symbol

Type: Type: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (Type)
x ∈ TVariables: Θ, B, Sch range over Type: the statement must hold for every choice. ((Θ B Sch : Type))
∀For all: The statement after the comma holds for every θ, b, s, s'. (∀ θ b s s', U θ b s = U θ b s')
NNatural numbers: The natural numbers 0, 1, 2, …. In Lean they start at 0. (ℕ)
x : TVariables: σ, σ' range over ℕ → Sch: the statement must hold for every choice. ((σ σ' : ℕ → Sch))
⟹Implies: If Sch, then Θ. (Between types, A → B is the type of functions from A to B.) (Sch → Θ)
rec: Nat.rec: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (Nat.rec)
f(x)Function application: U applied to θ, b, s. Lean writes f x; papers write f(x). (U θ b s)
=Equality: U(θ, b, s) and U(θ, b, s') are the same. (U θ b s = U θ b s')
motive: motive: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (motive)

theorem bz-schedule-independent-trajectory {Θ B Sch : Type} (U : Θ → B → Sch → Θ) (hU : ∀ θ b s s', U θ b s = U θ b s') (θ0 : Θ) (batch : ℕ → B) (σ σ' : ℕ → Sch) :
    ∀ T : ℕ, Nat.rec (motive := fun _ => Θ) θ0 (fun t θ => U θ (batch t) (σ t)) T = Nat.rec (motive := fun _ => Θ) θ0 (fun t θ => U θ (batch t) (σ' t)) T

Let U(θ, batch, schedule) be one training step. If U does not depend on the schedule (kernel grids, chunking, atomic order), then the parameters after every number of steps are identical under any per-step schedule sequence. Part of the plan 'Baguette-Z: closed-loop certified integer-state training'.

How it was proved

Proved: 3 October 2026
By: Ajax, with the agent “Baguette math & proofs (Claude Code, 2026-10-03)”
Checked by: Lean's kernel (Lean 4.33.1, with Mathlib); the proof is signed and replays offline
Axioms: None
Proof: 18 lines of Lean

If each update is schedule-independent, the whole training trajectory is

How it was proved

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