Encyclopedia Foundation Foundation Dalembert Necessity Gates Jcost Has Interaction
ARTICLE 3 claims 3 theorems
Foundation Dalembert Necessity Gates Jcost Has Interaction
A formal gate that separates the framework's cost function from a simpler quadratic alternative, by requiring that combining two comparisons is not merely additive.
The interaction requirement
The d'Alembert equation, named for Jean le Rond d'Alembert's 1747 work on vibrating strings, describes waves that propagate without changing shape. In the Recognition Science framework, a related structure governs how the cost of combining two comparisons behaves. The central object is the cost, a number assigned to each comparison that measures how expensive that recognition event is. The framework's key cost function is J(x) = (x + 1/x)/2 - 1, which the framework's machine-checked library of formal theorems proves is the unique function satisfying five plain conditions.
Those five conditions alone, however, do not force J. A counterexample in the framework's library shows that symmetry, normalization, twice-differentiability, calibration, and the existence of some combiner still allow a different branch: the quadratic-log function F(x) = (log x)^2 / 2, combined by P(u,v) = 2u + 2v. This function satisfies all five conditions but is not the framework's cost. The interaction gate closes that gap. It asserts that there exists at least one pair of positive numbers x and y such that the combined cost of the product and quotient is not purely additive in the two individual costs. In symbols: F(x·y) + F(x/y) ≠ 2F(x) + 2F(y).
The declaration Jcost_hasInteraction proves that the framework's cost J satisfies this gate. The proof exhibits a concrete witness: choosing x = 2 and y = 2 gives J(4) + J(1) = 9/8 + 0, while 2J(2) + 2J(2) = 1/4 + 1/4, and these are unequal. The quadratic-log function, by contrast, fails the gate: for every positive x and y, F(x·y) + F(x/y) equals 2F(x) + 2F(y) exactly. This is the theorem Fquad_noInteraction, which states that the quadratic branch has no interaction at all.
What the declaration does not claim is broader. It does not prove that J is unique; that uniqueness comes from the full forcing chain, not from this gate alone. It does not assert that the interaction gate is the only way to rule out the quadratic branch, only that it is the weakest such nondegeneracy condition the framework's library implements. And it does not say anything about whether the quadratic function is physically meaningful; the gate merely identifies it as outside the framework's forced structure.
THEOREM Jcost_hasInteraction · IndisputableMonolith/Foundation/DAlembert/NecessityGates.lean
theorem Jcost_hasInteraction : HasInteraction Jcost := by
-- Concrete witness: x = y = 2.
refine ⟨2, 2, by norm_num, by norm_num, ?_⟩
-- Evaluate numerically: J(4) + J(1) ≠ 2J(2) + 2J(2).
-- J(1)=0, J(2)=1/4, J(4)=9/8.
norm_num [Jcost]
THEOREM Jcost_hasInteraction · IndisputableMonolith/Foundation/DAlembert/NecessityGates.lean
theorem Jcost_hasInteraction : HasInteraction Jcost := by
-- Concrete witness: x = y = 2.
refine ⟨2, 2, by norm_num, by norm_num, ?_⟩
-- Evaluate numerically: J(4) + J(1) ≠ 2J(2) + 2J(2).
-- J(1)=0, J(2)=1/4, J(4)=9/8.
norm_num [Jcost]
THEOREM Fquad_noInteraction · IndisputableMonolith/Foundation/DAlembert/NecessityGates.lean
theorem Fquad_noInteraction : ¬ HasInteraction Fquad := by
intro h
rcases h with ⟨x, y, hx, hy, hneq⟩
exact hneq (Fquad_additive x y hx hy)
What this page does not claim
This declaration alone proves the uniqueness of J; uniqueness comes from the full forcing chain. The interaction gate is the only possible way to rule out the quadratic-log branch. The quadratic function is physically meaningful or meaningless; the gate only marks it as outside the framework.
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/DAlembert/NecessityGates.lean
expected axiom basis: [propext, Classical.choice, Quot.sound] (the Lean kernel's standard three; no RS-specific axioms)
A page whose claims cannot be reproduced this way does not ship. In production, every anchor links to the exact declaration in the public source release, and this block carries the build receipt for the page itself.
Derived articles
This page is generated by a question-recursion engine: the questions its answers raise become the next pages. The current agenda, with open targets marked red:
- How does the interaction gate relate to the full set of conditions that force J uniquely?
- What other nondegeneracy gates could replace or strengthen the interaction requirement?
- Does the quadratic-log function have any physical interpretation outside the framework?
MACHINE LAYER · GROUNDED CLAIM TABLE · CLICK TO EXPAND
THEOREM Jcost_hasInteraction · IndisputableMonolith/Foundation/DAlembert/NecessityGates.lean
theorem Jcost_hasInteraction : HasInteraction Jcost := by -- Concrete witness: x = y = 2. refine ⟨2, 2, by norm_num, by norm_num, ?_⟩ -- Evaluate numerically: J(4) + J(1) ≠ 2J(2) + 2J(2). -- J(1)=0, J(2)=1/4, J(4)=9/8. norm_num [Jcost]The declaration Jcost_hasInteraction proves that the framework's cost J satisfies this gate. Jcost_hasInteraction · IndisputableMonolith/Foundation/DAlembert/NecessityGates.leanTHEOREM Jcost_hasInteraction · IndisputableMonolith/Foundation/DAlembert/NecessityGates.lean
theorem Jcost_hasInteraction : HasInteraction Jcost := by -- Concrete witness: x = y = 2. refine ⟨2, 2, by norm_num, by norm_num, ?_⟩ -- Evaluate numerically: J(4) + J(1) ≠ 2J(2) + 2J(2). -- J(1)=0, J(2)=1/4, J(4)=9/8. norm_num [Jcost]The proof exhibits a concrete witness: choosing x = 2 and y = 2 gives J(4) + J(1) = 9/8 + 0, while 2J(2) + 2J(2) = 1/4 + 1/4, and these are unequal. Jcost_hasInteraction · IndisputableMonolith/Foundation/DAlembert/NecessityGates.leanTHEOREM Fquad_noInteraction · IndisputableMonolith/Foundation/DAlembert/NecessityGates.lean
theorem Fquad_noInteraction : ¬ HasInteraction Fquad := by intro h rcases h with ⟨x, y, hx, hy, hneq⟩ exact hneq (Fquad_additive x y hx hy)The quadratic-log function, by contrast, fails the gate: for every positive x and y, F(x·y) + F(x/y) equals 2F(x) + 2F(y) exactly. Fquad_noInteraction · IndisputableMonolith/Foundation/DAlembert/NecessityGates.lean