English
For x and u, the even-indexed position i+i of vecCons x u equals the corresponding Alt0-construction over vecAppend x u with itself.
Русский
Для x и u чётная позиция i+i vecCons x u равна соответствующей конструкции Alt0 над vecAppend x u.
LaTeX
$$$\\operatorname{cons\\_vecBit0\\_eq\\_alt0}(x,u,i) :$$
Lean4
/-- For an empty type, the recursion principle evaluates to whatever `congr`
makes of the base case.
-/
theorem recEmptyOption_of_card_eq_zero {P : Type u → Sort v} (finChoice : (n : ℕ) → Fin (n + 1))
(congr : {α β : Type u} → (_ : FinEnum α) → (_ : FinEnum β) → card β = card α → P α → P β)
(empty : P PEmpty.{u + 1}) (option : {α : Type u} → FinEnum α → P α → P (Option α)) (α : Type u) [FinEnum α]
(h : card α = 0) (_ : FinEnum PEmpty.{u + 1}) :
recEmptyOption finChoice congr empty option α = congr _ _ (h.trans card_eq_zero.symm) empty :=
by
unfold recEmptyOption
split
· congr 1; exact Subsingleton.allEq _ _
· exact Nat.noConfusion <| h.symm.trans ‹_›
