proofs
Proveddrift-lawdecodesnapshot-reuseattention-compute-law

Refresh law, one statement: per-token K/V traffic ≤ N(1 + (e−1)c) + n/T + N(e−1)c·T for refresh period T with 2cT ≤ 1

For all n,N,c∈R\htmlData{mx=1}{n, N, c \in \htmlData{mx=0}{\mathbb{R}}} and T∈N\htmlData{mx=1}{T \in \htmlData{mx=0}{\mathbb{N}}}, if 0<T0 \mathrel{\htmlData{mx=1}{<}} \htmlData{mx=0}{T}, 0≤c0 \mathrel{\htmlData{mx=1}{\leq}} \htmlData{mx=0}{c}, 0≤N0 \mathrel{\htmlData{mx=1}{\leq}} \htmlData{mx=0}{N} and 2c⋅T≤1\htmlData{mx=3}{\htmlData{mx=1}{2\htmlData{mx=0}{c}} \cdot \htmlData{mx=2}{T}} \mathrel{\htmlData{mx=4}{\leq}} 1, then

n+N⋅∑t=0T−1e2c⋅(t+1)T≤N⋅(1+(e1−1)⋅c)+(nT+N⋅(e1−1)⋅c⋅T)\begin{aligned} & \htmlData{mx=14}{\frac{\htmlData{mx=12}{\htmlData{mx=1}{n} + \htmlData{mx=11}{\htmlData{mx=2}{N} \cdot \htmlData{mx=10}{\sum_{t=0}^{\htmlData{mx=3}{T} - 1} \htmlData{mx=9}{e^{\htmlData{mx=8}{\htmlData{mx=5}{2\htmlData{mx=4}{c}} \cdot (\htmlData{mx=7}{\htmlData{mx=6}{t} + 1})}}}}}}}{\htmlData{mx=13}{T}}} \\ & \quad \htmlData{mx=0}{\leq} \htmlData{mx=34}{\htmlData{mx=21}{\htmlData{mx=15}{N} \cdot (\htmlData{mx=20}{1 + \htmlData{mx=19}{(\htmlData{mx=17}{\htmlData{mx=16}{e^{1}} - 1}) \cdot \htmlData{mx=18}{c}}})} + \left(\htmlData{mx=33}{\htmlData{mx=24}{\frac{\htmlData{mx=22}{n}}{\htmlData{mx=23}{T}}} + \htmlData{mx=32}{\htmlData{mx=30}{\htmlData{mx=28}{\htmlData{mx=25}{N} \cdot (\htmlData{mx=27}{\htmlData{mx=26}{e^{1}} - 1})} \cdot \htmlData{mx=29}{c}} \cdot \htmlData{mx=31}{T}}}\right)} \end{aligned}
What the notation means · or point at any symbol
RReal numbers
The real numbers: every point on the number line. (ℝ)
x ∈ TVariables
n, N, c range over real numbers: the statement must hold for every choice. ((n N c : ℝ))
NNatural numbers
The natural numbers 0, 1, 2, …. In Lean they start at 0. (ℕ)
<Less than
0 is strictly smaller than T. (0 < T)
≤At most
0 is less than or equal to c. (0 ≤ c)
·Multiplication
2 times c. (2 * c)
a/bDivision
n + N · (∑ t, exp(2 · c · (t + 1))) divided by T. In Lean, dividing by 0 gives 0. ((n + N * ∑ t ∈ Finset.range T, Real.exp…)
∑Summation
Add up exp(2 · c · (t + 1)) for t = 0, 1, …, T − 1. (∑ t ∈ Finset.range T, Real.exp (2 * c *…)
eˣExponential function
e raised to the power 2 · c · (t + 1), where e ≈ 2.718. (Real.exp (2 * c * (t + 1)))
+Addition
t plus 1. (t + 1)
-Subtraction
exp(1) minus 1. On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. (Real.exp 1 - 1)

Assembled bound for exact snapshot reuse under linear drift c. Loop 2 of the Drift Law shows the drift budget 2cT ≤ 1 caps T at 1/(2c), so the √n optimum is only a transient regime (n ≲ N/(4(e−1)c)); the asymptotic static-refresh traffic is Θ(c n / log n), and only the bounded-drift persistent snapshot is flat in n. Part of 'The Drift Law: decode K/V traffic is set by how far the query drifts' (exact snapshot reuse at decode).

How it was proved

Proved
5 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
Classical.choice, Quot.sound, propext
Proof
45 lines of Lean
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