Encyclopedia Foundation Foundation Recognition Signature Gauge Scalar Cost Kernel Eq Signature Of Comple
ARTICLE 4 claims 3 theorems 1 model
Foundation Recognition Signature Gauge Scalar Cost Kernel Eq Signature Of Comple
A single number can summarize a state only when the measurement system is rich enough to tell every distinct state apart.
The completeness condition
A recognition event, in the Recognition Science framework, is a discrete act of measurement: an observer looks at a state and records a value. The framework's library, a machine-checked collection of formal results, asks when such a record is enough. The declaration named scalar_cost_kernel_eq_signature_of_complete answers a precise version of that question. It proves that if a scalar cost function, a single real number assigned to each state, is complete for a family of observables, then two states have the same cost exactly when they have the same full recognition signature. The full recognition signature is the tuple of values that every observable in the family returns for that state.
The condition that makes the declaration true is called ScalarCostCompleteFor. It states, by definition, that cost x = cost y holds if and only if the two states are indistinguishable by the entire family of observables. The declaration then simply restates that definition as an equivalence: under the completeness hypothesis, the scalar cost kernel, the set of pairs with equal cost, coincides with the signature kernel, the set of pairs with identical full signatures. This is a result in the machine-checked library, proved with no unproved assumptions beyond the standard logical axioms.
The declaration's force is negative as much as positive. The library also proves that a single Boolean coordinate, a one-bit measurement, is not complete in general. For the state space Bool × Bool, the first bit alone cannot separate (false, false) from (false, true); both states share the same first bit and therefore the same scalar cost, yet they are distinct states. The two-coordinate family, reading both bits, does separate them. The completeness hypothesis is therefore not automatic. It is a real condition that must be established for each measurement system before the scalar cost can stand in for the full signature.
In Recognition Science, this result sharpens the role of the scalar cost in the forcing chain. The cost function J(x) = (x + 1/x)/2 - 1 is proved unique under its own axioms, but this declaration does not say that any scalar cost is complete. It says that a scalar cost is a faithful summary only when the observable family is rich enough. The single number is a gauge, a coordinate on the quotient of states by indistinguishability, and the completeness condition is what licenses reading the number as the whole story.
THEOREM scalar_cost_kernel_eq_signature_of_complete · IndisputableMonolith/Foundation/RecognitionSignatureGauge.lean
/-- Under the explicit completeness hypothesis, the scalar-cost kernel matches
the full recognition-signature equivalence. Without this hypothesis, scalar
cost equality is only a cost observable. -/
theorem scalar_cost_kernel_eq_signature_of_complete
(F : Set (X → C)) (cost : X → ℝ) (hcomplete : ScalarCostCompleteFor F cost)
(x y : X) :
cost x = cost y ↔ SameRecognitionSignature F x y :=
hcomplete x y
MODEL ScalarCostCompleteFor · IndisputableMonolith/Foundation/RecognitionSignatureGauge.lean
/-- A scalar cost is complete for a recognition family only if equality of the
scalar is equivalent to equality of the full recognition signature. This is a
separate hypothesis, not a consequence of having a scalar cost. -/
def ScalarCostCompleteFor (F : Set (X → C)) (cost : X → ℝ) : Prop :=
∀ x y : X, cost x = cost y ↔ SameRecognitionSignature F x y
THEOREM one_boolean_coordinate_not_complete · IndisputableMonolith/Foundation/RecognitionSignatureGauge.lean
/-- One Boolean coordinate fails to separate the two states with the same first
bit and different second bit. -/
theorem one_boolean_coordinate_not_complete :
∃ x y : PairBoolState,
x ≠ y ∧ SameRecognitionSignature firstBitFamily x y := by
refine ⟨(false, false), (false, true), ?_, ?_⟩
· decide
· intro f hf
have hf' : f = firstBit := hf
rw [hf']
rfl
THEOREM pairBitFamily_separating · IndisputableMonolith/Foundation/RecognitionSignatureGauge.lean
/-- The two coordinate Boolean recognizers separate all states of `Bool × Bool`. -/
theorem pairBitFamily_separating :
∀ x y : PairBoolState, SameRecognitionSignature pairBitFamily x y → x = y := by
intro x y hsig
cases x with
| mk x₁ x₂ =>
cases y with
| mk y₁ y₂ =>
have h₁ : x₁ = y₁ := by
exact hsig firstBit (Or.inl rfl)
have h₂ : x₂ = y₂ := by
exact hsig secondBit (Or.inr rfl)
cases h₁
cases h₂
rfl
What this page does not claim
The declaration does not claim that any scalar cost function is complete for an arbitrary family of observables. The declaration does not claim that a single Boolean coordinate is sufficient to distinguish all states in every state space. The declaration does not claim that the cost function J itself is complete for any particular recognition system.
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/RecognitionSignatureGauge.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:
- What is the precise definition of the full recognition signature for a given family of observables?
- Under what conditions does a family of observables separate all states in a given state space?
- How does the completeness condition for scalar cost relate to the uniqueness theorem for the cost function J?
- What role does the PhysicalQuotient play in the statement that observables descend to the quotient?
MACHINE LAYER · GROUNDED CLAIM TABLE · CLICK TO EXPAND
THEOREM scalar_cost_kernel_eq_signature_of_complete · IndisputableMonolith/Foundation/RecognitionSignatureGauge.lean
/-- Under the explicit completeness hypothesis, the scalar-cost kernel matches the full recognition-signature equivalence. Without this hypothesis, scalar cost equality is only a cost observable. -/ theorem scalar_cost_kernel_eq_signature_of_complete (F : Set (X → C)) (cost : X → ℝ) (hcomplete : ScalarCostCompleteFor F cost) (x y : X) : cost x = cost y ↔ SameRecognitionSignature F x y := hcomplete x yIt proves that if a scalar cost function is complete for a family of observables, then two states have the same cost exactly when they have the same full recognition signature. scalar_cost_kernel_eq_signature_of_complete · IndisputableMonolith/Foundation/RecognitionSignatureGauge.leanMODEL ScalarCostCompleteFor · IndisputableMonolith/Foundation/RecognitionSignatureGauge.lean
/-- A scalar cost is complete for a recognition family only if equality of the scalar is equivalent to equality of the full recognition signature. This is a separate hypothesis, not a consequence of having a scalar cost. -/ def ScalarCostCompleteFor (F : Set (X → C)) (cost : X → ℝ) : Prop := ∀ x y : X, cost x = cost y ↔ SameRecognitionSignature F x yThe condition that makes the declaration true is called ScalarCostCompleteFor. ScalarCostCompleteFor · IndisputableMonolith/Foundation/RecognitionSignatureGauge.leanTHEOREM one_boolean_coordinate_not_complete · IndisputableMonolith/Foundation/RecognitionSignatureGauge.lean
/-- One Boolean coordinate fails to separate the two states with the same first bit and different second bit. -/ theorem one_boolean_coordinate_not_complete : ∃ x y : PairBoolState, x ≠ y ∧ SameRecognitionSignature firstBitFamily x y := by refine ⟨(false, false), (false, true), ?_, ?_⟩ · decide · intro f hf have hf' : f = firstBit := hf rw [hf'] rflThe library also proves that a single Boolean coordinate is not complete in general. one_boolean_coordinate_not_complete · IndisputableMonolith/Foundation/RecognitionSignatureGauge.leanTHEOREM pairBitFamily_separating · IndisputableMonolith/Foundation/RecognitionSignatureGauge.lean
/-- The two coordinate Boolean recognizers separate all states of `Bool × Bool`. -/ theorem pairBitFamily_separating : ∀ x y : PairBoolState, SameRecognitionSignature pairBitFamily x y → x = y := by intro x y hsig cases x with | mk x₁ x₂ => cases y with | mk y₁ y₂ => have h₁ : x₁ = y₁ := by exact hsig firstBit (Or.inl rfl) have h₂ : x₂ = y₂ := by exact hsig secondBit (Or.inr rfl) cases h₁ cases h₂ rflThe two-coordinate family, reading both bits, does separate them. pairBitFamily_separating · IndisputableMonolith/Foundation/RecognitionSignatureGauge.lean