English
For all m,n in PosNum, m + succ n = succ (m + n).
Русский
Для любых m,n из PosNum справедливо m + succ n = succ (m + n).
LaTeX
$$$\\forall m,n:\\mathrm{PosNum},\\; m + \\mathrm{succ} n = \\mathrm{succ}(m + n)$$$
Lean4
theorem add_succ : ∀ m n : PosNum, m + succ n = succ (m + n)
| 1, b => by simp [one_add]
| bit0 a, 1 => congr_arg bit0 (add_one a)
| bit1 a, 1 => congr_arg bit1 (add_one a)
| bit0 _, bit0 _ => rfl
| bit0 a, bit1 b => congr_arg bit0 (add_succ a b)
| bit1 _, bit0 _ => rfl
| bit1 a, bit1 b => congr_arg bit1 (add_succ a b)