Sect. 3.1, p. 80: and once imply strictly increasing
ProvedHackCosmologicalAQFT.strictMonoOn_of_first_friedmannLet and let be an open, connected set of cosmological times. Let be twice continuously differentiable on with there, and suppose the first Friedmann equation holds on with
If at a single instant , then is strictly increasing on all of .
This is the remark in Hack (2016), p. 80: "if for all times and at one instant of time, then will be strictly increasing for all times." It underlies the use of and the redshift as time variables.
import Mathlib import Definitions.Def_hack_flrw_friedmann
namespace HackCosmologicalAQFT
/-- Hack 2016, Sect. 3.1, p. 80: if `ρ > 0` at all times and `ȧ > 0` at one instant, the first
Friedmann equation forces `a` to be strictly increasing at all times. -/
theorem strictMonoOn_of_first_friedmann (G : ℝ) (hG : 0 < G) (a ρ : ℝ → ℝ) (I : Set ℝ)
(hI : IsOpen I) (hIc : IsPreconnected I)
(ha : ContDiffOn ℝ 2 a I) (hpos : ∀ t ∈ I, 0 < a t)
(hF1 : FirstFriedmannEq G a ρ I) (hρ : ∀ t ∈ I, 0 < ρ t)
(t₁ : ℝ) (ht₁ : t₁ ∈ I) (hexp : 0 < deriv a t₁) :
StrictMonoOn a I := by sorry
end HackCosmologicalAQFTRead-back
What the Lean code literally says, in plain math · Aristotle (Harmonic) — non-blind, same agent as drafter
NON-BLIND READ-BACK — not independent testimony. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statement, with full knowledge of the source text and of the intended meaning. It was not produced by a blind, independent auditor, so it may silently read the intended meaning into the code. Reviewers must not mistake it for independent testimony; compare the Lean code against the source directly.
Fix with , functions , and a set . Assume:
- is open and preconnected;
- is twice continuously differentiable on , and for all ;
- for all : ;
- for all ;
- there is a point with .
Conclusion. is strictly increasing on : for all with we have . Assumption 5 forces to be nonempty. No hypothesis on the second Friedmann equation or on a pressure is made.