English
If f is surjective, then g1 ∘ f = g2 ∘ f iff g1 = g2 for BiheytingHom.
Русский
Если f сюръективен, то g1 ∘ f = g2 ∘ f тогда и только тогда, когда g1 = g2 для BiheytingHom.
LaTeX
$$$\forall f:\mathrm{BiheytingHom}(\alpha,\beta),\ (\mathrm{Surjective}(f))\to( g_1\circ f = g_2\circ f \iff g_1 = g_2 )$$$
Lean4
@[simp]
theorem cancel_left (hg : Injective g) : g.comp f₁ = g.comp f₂ ↔ f₁ = f₂ :=
⟨fun h => BiheytingHom.ext fun a => hg <| by rw [← comp_apply, h, comp_apply], congr_arg _⟩