finger_is_10
plain-language theorem explainer
Finger enumeration satisfies the ten-fold cardinality condition in the cross-domain module. Researchers cataloging 2·5 factorizations across biology and structure would cite this result. The proof reduces the claim to a decidable equality on the finite type by unfolding the predicate.
Claim. Let $F$ be the type consisting of ten fingers (five left-hand and five right-hand variants). Then the cardinality of $F$ equals 10.
background
The TenFoldCombinations module documents structural instances where domains enumerate to 10 = 2·5, including fingers, digits, and d-block elements. HasTenFold is the predicate asserting that a finite type has cardinality exactly 10. Finger is the inductive type with ten constructors for the paired fingers. This setting supports the claim that the doubling arises from pairing a 5-element structure with an orientation or polarity, as stated in the module documentation.
proof idea
The proof is a one-line wrapper. It unfolds HasTenFold to the equality Fintype.card Finger = 10 and invokes the decide tactic, which succeeds on the explicitly finite inductive type.
why it matters
The result supplies the finger instance to the tenFoldCombinationsCert certificate that aggregates multiple ten-fold verifications. It realizes the C17 claim of ten-fold combinations in the module, linking to the universal factorization 10 = 2·D. Within Recognition Science this instance illustrates the cross-domain reach of the 2·5 pattern without invoking the phi-ladder or forcing chain directly.
Switch to Lean above to see the machine-checked source, dependencies, and usage graph.