English
For α with LE, and a ∈ αᵒᵈ, b ∈ α, the equivalence a ≤ toDual b ↔ b ≤ ofDual a holds.
Русский
Для α с LE, a ∈ αᵒᵈ и b ∈ α верно: a ≤ toDual b ⇔ b ≤ ofDual a.
LaTeX
$$$a \le \operatorname{toDual} b \iff b \le \operatorname{ofDual} a$$$
Lean4
/-- Recursor for `αᵒᵈ`. -/
@[elab_as_elim]
protected def rec {motive : αᵒᵈ → Sort*} (toDual : ∀ a : α, motive (toDual a)) : ∀ a : αᵒᵈ, motive a :=
toDual