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Fig. 1: equal entropies at T=0T=0T=0 — no finite number of cooling steps reaches absolute zero

Proved
ThirdLawThermo.demagnetization_cycles_never_reach_zero

by Lucas · Oct 9, 2026 · Mathlib 0df444a (Lean v4.33.1)

mathematical-physicsstatistical-mechanicsthermodynamicsthird-law

Let S(T,x)S(T,x)S(T,x) be the equilibrium entropy at temperature T≥0T\ge0T≥0 and parameter value xxx, and fix two parameter values X1,X2X_1,X_2X1​,X2​ (in magnetic refrigeration, a lower and a higher magnetic field). One cooling step at temperature TnT_nTn​ consists of an isothermal change of the parameter from X1X_1X1​ to X2X_2X2​ at temperature TnT_nTn​, followed by an isentropic (reversible adiabatic) change back from X2X_2X2​ to X1X_1X1​, which ends at the temperature Tn+1≥0T_{n+1}\ge0Tn+1​≥0 determined by

S(Tn+1,X1)=S(Tn,X2).S(T_{n+1},X_1)=S(T_n,X_2).S(Tn+1​,X1​)=S(Tn​,X2​).

Assume that T↦S(T,X2)T\mapsto S(T,X_2)T↦S(T,X2​) is strictly increasing on [0,∞)[0,\infty)[0,∞) and that there is no entropy difference at absolute zero, S(0,X1)=S(0,X2)S(0,X_1)=S(0,X_2)S(0,X1​)=S(0,X2​).

Theorem. Starting from T0>0T_0>0T0​>0, every finite sequence of mmm such cooling steps ends at a positive temperature: Tm>0T_m>0Tm​>0. Hence absolute zero cannot be reached in a finite number of steps.

This is the right-hand panel of Fig. 1 of the source, the mechanism by which the third law implies the unattainability of absolute zero.

Formalization Note The temperatures are given as a sequence T:N→RT:\mathbb N\to\mathbb RT:N→R of which only T0,…,TmT_0,\dots,T_mT0​,…,Tm​ are constrained. The converse situation of the left-hand panel (S(0,X1)≠S(0,X2)S(0,X_1)\neq S(0,X_2)S(0,X1​)=S(0,X2​)) is not part of this statement.

Preamble
import Definitions.Def_ThirdLawThermo_Defs
import Mathlib

open Filter Topology
Formal statement
namespace ThirdLawThermo

theorem demagnetization_cycles_never_reach_zero {X : Type*} (S : ℝ → X → ℝ) (x₁ x₂ : X)
    (hmono₂ : StrictMonoOn (fun T => S T x₂) (Set.Ici 0))
    (hNernst : S 0 x₁ = S 0 x₂)
    (T : ℕ → ℝ) (m : ℕ) (hT₀ : 0 < T 0)
    (hstep : ∀ n < m, 0 ≤ T (n + 1) ∧ S (T (n + 1)) x₁ = S (T n) x₂) :
    0 < T m := by sorry

end ThirdLawThermo
Source
Wikipedia, "Third law of thermodynamics", revision oldid=1369622029, https://en.wikipedia.org/w/index.php?title=Third_law_of_thermodynamics&oldid=1369622029; section "Consequences", subsection "Absolute zero" and Fig. 1 ("Suppose that the temperature of a substance can be reduced in an isentropic process by changing the parameter X from X2 to X1 ... If there were an entropy difference at absolute zero, T = 0 could be reached in a finite number of steps. However, at T = 0 there is no entropy difference, so an infinite number of steps would be needed"; caption: "Right: An infinite number of steps is needed since S(0, X1) = S(0, X2)"); subsection "Example: magnetic refrigeration" for the two-step cycle.
Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic), the drafting agent itself; non-blind read-back

Disclosure: this read-back is NOT blind. It was written by the same agent (Aristotle, by Harmonic) that drafted these Lean statements, with full knowledge of the source text and of the intended meaning. It is therefore not independent testimony and must not be mistaken for an independent audit. Reviewers should check the Lean code against the source themselves.

Let XXX be an arbitrary type, S:R×X→RS:\mathbb R\times X\to\mathbb RS:R×X→R an arbitrary function written S(T,x)S(T,x)S(T,x), and x1,x2∈Xx_1,x_2\in Xx1​,x2​∈X (not required to be distinct). Assume

  1. T↦S(T,x2)T\mapsto S(T,x_2)T↦S(T,x2​) is strictly increasing on [0,∞)[0,\infty)[0,∞) (no monotonicity is assumed in the variable TTT for x1x_1x1​);
  2. S(0,x1)=S(0,x2)S(0,x_1)=S(0,x_2)S(0,x1​)=S(0,x2​).

Let T0,T1,T2,…T_0,T_1,T_2,\dotsT0​,T1​,T2​,… be any real sequence and mmm any natural number, with T0>0T_0>0T0​>0, and assume that for every n<mn<mn<m:

Tn+1≥0andS(Tn+1,x1)=S(Tn,x2).T_{n+1}\ge0\quad\text{and}\quad S(T_{n+1},x_1)=S(T_n,x_2).Tn+1​≥0andS(Tn+1​,x1​)=S(Tn​,x2​).

Then Tm>0T_m>0Tm​>0. For m=0m=0m=0 the claim is just T0>0T_0>0T0​>0. Terms TnT_nTn​ with n>mn>mn>m are unconstrained.

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