Foundation Primitive Recognition Calculus Prcmonotone Dalembert D Alembert Ge On

The order-only floor

The d'Alembert equation, H(s+t) + H(s−t) = 2 H(s) H(t), is a classical functional equation whose solutions include the hyperbolic cosine and cosine families. The declaration dAlembert_ge_one_of_monotone proves a plain order-theoretic fact about such solutions: if H(0) = 1 and H is nondecreasing on the nonnegative reals, then H(t) ≥ 1 for every t ≥ 0. Monotonicity alone, applied from the base value 1, forces the floor in one step. No continuity, no smoothness, and no least-upper-bound axiom enter the proof.

The statement matters because it excludes the bounded cosine branch, H(t) = cos(c·t), which dips below 1 and therefore cannot be monotone on [0, ∞). This leaves the unbounded cosh branch as the only order-compatible family. In the Recognition Science framework, this is the first place the monotone route does real work that continuity used to do: it pins the sign of the square root in the difference law, resolving the sign crux by order alone. The framework's library of machine-checked theorems uses this brick to show that the cost form is forced by monotonicity instead of continuity, a completeness-free substitute available on any Archimedean ordered field.

The declaration does not claim that every solution of the d'Alembert equation is monotone, nor that monotonicity alone forces the full cosh form without the other hypotheses (evenness, normalization, and the composition law). It establishes only the floor H ≥ 1 on the nonnegative reals, given H(0) = 1 and monotonicity. The sign-fixing and the final cosh classification are separate theorems in the same block.

/-- **§9 order-only constraint 2 (completeness-free): the cosh floor `H ≥ 1`.**
A d'Alembert solution that is monotone on `[0,∞)` with `H 0 = 1` stays `≥ 1`
there. The floor is forced by order alone: monotonicity from the base value `1`
gives it in one step. This excludes the bounded "cosine" branch `H = cos(c·)` of
d'Alembert (which dips below `1`) using no analytic input, isolating the
unbounded cosh branch as the only order-compatible family — the first place the
§9 monotone route does real work that continuity used to do. -/
theorem dAlembert_ge_one_of_monotone {H : ℝ → ℝ}
    (h0 : H 0 = 1) (hmono : MonotoneOn H (Set.Ici (0 : ℝ))) :
    ∀ t, 0 ≤ t → 1 ≤ H t := by
  intro t ht
  have hle := hmono Set.left_mem_Ici (Set.mem_Ici.mpr ht) ht
  rwa [h0] at hle

/-- **§9 order-only constraint 2 (completeness-free): the cosh floor `H ≥ 1`.**
A d'Alembert solution that is monotone on `[0,∞)` with `H 0 = 1` stays `≥ 1`
there. The floor is forced by order alone: monotonicity from the base value `1`
gives it in one step. This excludes the bounded "cosine" branch `H = cos(c·)` of
d'Alembert (which dips below `1`) using no analytic input, isolating the
unbounded cosh branch as the only order-compatible family — the first place the
§9 monotone route does real work that continuity used to do. -/
theorem dAlembert_ge_one_of_monotone {H : ℝ → ℝ}
    (h0 : H 0 = 1) (hmono : MonotoneOn H (Set.Ici (0 : ℝ))) :
    ∀ t, 0 ≤ t → 1 ≤ H t := by
  intro t ht
  have hle := hmono Set.left_mem_Ici (Set.mem_Ici.mpr ht) ht
  rwa [h0] at hle

/-- **§9 order-only constraint 2 (completeness-free): the cosh floor `H ≥ 1`.**
A d'Alembert solution that is monotone on `[0,∞)` with `H 0 = 1` stays `≥ 1`
there. The floor is forced by order alone: monotonicity from the base value `1`
gives it in one step. This excludes the bounded "cosine" branch `H = cos(c·)` of
d'Alembert (which dips below `1`) using no analytic input, isolating the
unbounded cosh branch as the only order-compatible family — the first place the
§9 monotone route does real work that continuity used to do. -/
theorem dAlembert_ge_one_of_monotone {H : ℝ → ℝ}
    (h0 : H 0 = 1) (hmono : MonotoneOn H (Set.Ici (0 : ℝ))) :
    ∀ t, 0 ≤ t → 1 ≤ H t := by
  intro t ht
  have hle := hmono Set.left_mem_Ici (Set.mem_Ici.mpr ht) ht
  rwa [h0] at hle

What this page does not claim

The declaration does not claim that every d'Alembert solution is monotone. It does not claim that monotonicity alone forces the cosh form without evenness, normalization, and the composition law. It does not claim to resolve the sign of the square root; that is a separate theorem.

Verify this page

Every tagged claim above names its theorem. To check one yourself rather than trust this page, elaborate the source module with Lean 4 and audit its axiom basis:

$ lake env lean IndisputableMonolith/Foundation/PrimitiveRecognitionCalculus/PRCMonotoneDAlembert.lean
expected axiom basis: [propext, Classical.choice, Quot.sound] (the Lean kernel's standard three; no RS-specific axioms)

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