English
For any a,b in Sum.Lex, a < b is equivalent to the Lex LT comparison of their images under ofLex.
Русский
Для любых a,b из Sum.Lex верно: a < b эквивалентно LT-сравнению образов ofLex a и ofLex b.
LaTeX
$$$a < b \iff \text{Sum.Lex.LT.lt} (\mathrm{ofLex}\; a) (\mathrm{ofLex}\; b).$$$
Lean4
theorem lt_def [LT α] [LT β] {a b : α ⊕ₗ β} : a < b ↔ Lex (· < ·) (· < ·) (ofLex a) (ofLex b) :=
Iff.rfl