English
For any l and n, applying lower' to l with n and then raise' to the result recovers the original list: raise'(lower'(l,n), n) = l.
Русский
Для любых l и n применение lower' к l с параметром n, затем raise' к результату возвращает исходный список: raise'(lower'(l,n), n) = l.
LaTeX
$$$\forall l\,n,\, \text{raise'}(\text{lower'}(l,n), n) = l$$$
Lean4
theorem lower_raise' : ∀ l n, lower' (raise' l n) n = l
| [], _ => rfl
| m :: l, n => by simp [raise', lower', lower_raise']