both_equal_5_at_3
The declaration shows that the Fibonacci deficit and integration measure routes both yield coherence exponent 5 precisely when the spatial dimension is 3. Researchers tracing the Recognition Science forcing chain would cite this result when confirming that only D=3 satisfies the dual constraints on the exponent. The proof is a direct computation via the decide tactic on the explicit definitions of the two expressions.
claimLet $k_{fib}(D) := 2^D - D$ and $k_{int}(D) := D + 2$. Then $k_{fib}(3) = 5$ and $k_{int}(3) = 5$.
background
In the Coherence Exponent Uniqueness module, two independent expressions are defined for the coherence exponent k. The Fibonacci deficit route sets k_fib(D) = 2^D - D, which at D=3 yields 5, matching the fifth Fibonacci number. The integration measure route sets k_int(D) = D + 2, which also equals 5 at D=3.
proof idea
The proof is a one-line wrapper that invokes the decide tactic to verify the equality after substituting the explicit definitions of k_fib and k_int at the natural number 3.
why it matters in Recognition Science
This result supplies the agreement clause in the coherenceExponentCert certificate, which collects the agreements and disagreements to certify that k=5 is forced uniquely at D=3. It directly supports the master theorem exponent_unique_at_D3 in the module. Within the Recognition Science framework, it confirms the eight-tick octave and D=3 spatial dimensions by showing the exponent matches the required value only in three dimensions.
scope and limits
- Does not establish the physical interpretation of the coherence exponent.
- Does not derive the expressions k_fib or k_int from more primitive axioms.
- Does not address continuous dimensions or non-integer cases.
- Does not prove that k=5 is the only possible exponent outside these two routes.
Lean usage
def coherenceExponentCert : CoherenceExponentCert where agree_at_3 := agreement_at_3 both_five := both_equal_5_at_3 disagree_1 := disagreement_at_1 disagree_2 := disagreement_at_2 disagree_4 := disagreement_at_4formal statement (Lean)
46theorem both_equal_5_at_3 : k_fib 3 = 5 ∧ k_int 3 = 5 := by decide
proof body
Term-mode proof.
47
48/-- Disagreement at D = 1. -/