English
For all m,n,k, psub m n = some k iff k+n = m.
Русский
Для всех m,n,k: psub m n = some k ⇔ k+n = m.
LaTeX
$$$$ \forall {m n k}: \mathrm{Nat}, \ psub(m,n) = \mathrm{some} k \iff k+n = m. $$$$
Lean4
theorem psub_eq_some {m : ℕ} : ∀ {n k}, psub m n = some k ↔ k + n = m
| 0, k => by simp [eq_comm]
| n + 1, k => by
apply Option.bind_eq_some_iff.trans
simp only [psub_eq_some, ppred_eq_some]
simp [add_comm, add_left_comm]