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Linear bound is dominated by a polynomial bound

Proved
CookLevin.polyBound_of_linear

by Mazecto · Sep 21, 2026 · Mathlib 0df444a (Lean v4.33.1)

cost-modellinear-timepolynomial-time

Every linear step bound of the form c0(n+1)c_0 (n + 1)c0​(n+1) is bounded by a canonical polynomial bound polyBound(c,d,n)=c(n+1)d\mathrm{polyBound}(c, d, n) = c (n + 1)^dpolyBound(c,d,n)=c(n+1)d.

Specifically, for any coefficient c0∈Nc_0 \in \mathbb{N}c0​∈N, choosing c=c0c = c_0c=c0​ and degree d=1d = 1d=1 yields:

c0(n+1)≤polyBound(c0,1,n)=c0(n+1)1=c0(n+1)c_0 (n + 1) \le \mathrm{polyBound}(c_0, 1, n) = c_0 (n + 1)^1 = c_0 (n + 1)c0​(n+1)≤polyBound(c0​,1,n)=c0​(n+1)1=c0​(n+1)

which holds with equality for all n∈Nn \in \mathbb{N}n∈N.

This lemma bridges linear-time machine algorithms into the formal polynomial-time complexity framework PolyTimeDecidable.

Preamble
import Definitions.Def_CookLevin_Cost
Formal statement
namespace CookLevin
theorem polyBound_of_linear (c0 : Nat) :
    ∃ c d : Nat, ∀ n : Nat, c0 * (n + 1) ≤ polyBound c d n := by sorry
end CookLevin
Source
https://github.com/Rizvonium/cook_levin_lean_v1/blob/main/CookLevinLean/Cost.lean#L109

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