English
If g is injective, then g ∘ f1 = g ∘ f2 implies f1 = f2.
Русский
Если g инъективен, то g ∘ f1 = g ∘ f2 предполагает f1 = f2.
LaTeX
$$$\forall f_1,f_2:\mathrm{HeytingHom}(\alpha,\beta),\ g:\mathrm{HeytingHom}(\beta,\gamma),\ (\mathrm{Injective}(g))\to\Bigl( g\circ f_1 = g\circ f_2 \iff f_1 = f_2 \Bigr)$$$
Lean4
@[simp]
theorem cancel_left (hg : Injective g) : g.comp f₁ = g.comp f₂ ↔ f₁ = f₂ :=
⟨fun h => HeytingHom.ext fun a => hg <| by rw [← comp_apply, h, comp_apply], congr_arg _⟩