English
vecAlt1 distributes over vecCons similarly to Alt0 with the right offset.
Русский
vecAlt1 распределяется по vecCons аналогично Alt0, с учётом смещения.
LaTeX
$$$\\text{vecAlt1}(h)(\\text{vecCons }x u) = \\text{vecCons }y (\\text{vecAlt1}(by cutsat) \\, u)$$$
Lean4
@[simp]
theorem cons_vecAlt1 (h : m + 1 + 1 = n + 1 + (n + 1)) (x y : α) (u : Fin m → α) :
vecAlt1 h (vecCons x (vecCons y u)) = vecCons y (vecAlt1 (by cutsat) u) :=
by
ext i
simp_rw [vecAlt1]
rcases i with ⟨⟨⟩ | i, hi⟩
· rfl
· simp [vecAlt1, Nat.add_right_comm, ← Nat.add_assoc]