Encyclopedia Foundation Foundation Pinch Algebra Finite Operations From Budget
ARTICLE 2 claims 2 theorems
Foundation Pinch Algebra Finite Operations From Budget
A simple theorem about dividing a budget by a cost per operation proves that any finite resource can only buy finitely many steps.
The finite budget theorem
A budget is a finite amount of something, and an operation has a cost. If each operation costs a positive amount, and the budget is not negative, then the number of operations you can afford is bounded by the budget divided by the cost. The Recognition Science declaration finite_operations_from_budget formalizes this: if the number of operations times the cost fits within the budget, then that number is at most the budget divided by the cost.
This is a theorem in the machine-checked library of formal theorems, proved from the axioms of arithmetic. It is not a statement about physics or about the framework's own ledger of recognition events. It is a general mathematical fact: a finite resource cannot buy an infinite sequence of positive-cost steps. The theorem holds for any real numbers representing cost and budget, as long as the cost is positive and the budget is nonnegative.
The proof is a direct manipulation of inequalities. If n * cost ≤ budget and cost > 0, then dividing both sides by cost gives n ≤ budget / cost. The declaration simply records this step. It does not say what the operations are, what the budget is, or why the cost is positive. It is a tool, not a physical law.
In the Recognition Science framework, this theorem is a finite-capacity veto in its simplest form. It is cited as the basic set-theoretic obstruction behind the idea that a finite ledger cannot support an infinite process. But the theorem itself is neutral: it applies to any setting where costs are positive and budgets are finite, not only to recognition events.
THEOREM finite_operations_from_budget · IndisputableMonolith/Foundation/PinchAlgebra.lean
/-- **F5.3.1/3.2**: If the cost per operation is positive and the budget is finite,
only finitely many operations can be performed. -/
theorem finite_operations_from_budget {n : ℕ} {cost budget : ℝ}
(hcost : 0 < cost) (hbudget : 0 ≤ budget)
(hfit : n * cost ≤ budget) :
(n : ℝ) ≤ budget / cost := by
rwa [le_div_iff₀ hcost]
THEOREM finite_not_onto_infinite · IndisputableMonolith/Foundation/PinchAlgebra.lean
/-- **F5.3.2**: A function that is not surjective cannot map finite sets
onto infinite sets. (Basic set-theoretic obstruction.)
This is the finite-capacity veto in its simplest form. -/
theorem finite_not_onto_infinite {α β : Type*} (f : α → β)
[Finite α] (hβ : Infinite β) : ¬Function.Surjective f := by
intro hsurj
have : Finite β := Finite.of_surjective f hsurj
exact not_finite β
What this page does not claim
The theorem does not specify what operations or budgets are. It does not prove that any particular process in the framework is finite. It does not depend on any Recognition Science axioms.
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/PinchAlgebra.lean
expected axiom basis: [propext, Classical.choice, Quot.sound] (the Lean kernel's standard three; no RS-specific axioms)
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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:
- What counts as an operation in the Recognition Science framework?
- How does the finite-capacity veto connect to the forcing chain of theorems?
- What is the role of this theorem in the BSD Stage 6 template?
MACHINE LAYER · GROUNDED CLAIM TABLE · CLICK TO EXPAND
THEOREM finite_operations_from_budget · IndisputableMonolith/Foundation/PinchAlgebra.lean
/-- **F5.3.1/3.2**: If the cost per operation is positive and the budget is finite, only finitely many operations can be performed. -/ theorem finite_operations_from_budget {n : ℕ} {cost budget : ℝ} (hcost : 0 < cost) (hbudget : 0 ≤ budget) (hfit : n * cost ≤ budget) : (n : ℝ) ≤ budget / cost := by rwa [le_div_iff₀ hcost]If each operation costs a positive amount, and the budget is not negative, then the number of operations you can afford is bounded by the budget divided by the cost. finite_operations_from_budget · IndisputableMonolith/Foundation/PinchAlgebra.leanTHEOREM finite_not_onto_infinite · IndisputableMonolith/Foundation/PinchAlgebra.lean
/-- **F5.3.2**: A function that is not surjective cannot map finite sets onto infinite sets. (Basic set-theoretic obstruction.) This is the finite-capacity veto in its simplest form. -/ theorem finite_not_onto_infinite {α β : Type*} (f : α → β) [Finite α] (hβ : Infinite β) : ¬Function.Surjective f := by intro hsurj have : Finite β := Finite.of_surjective f hsurj exact not_finite βA finite resource cannot buy an infinite sequence of positive-cost steps. finite_not_onto_infinite · IndisputableMonolith/Foundation/PinchAlgebra.lean