English
Let p₁ be a predicate on α, s₁ a Setoid on α, s₂ a Setoid on Subtype p₁, p₂ a predicate on Quotient s₁, hp₂ a compatibility between p₁ and p₂, and h a compatibility of the relations. Then there is an equivalence between the subtype {x ∈ α | p₂ x} and the quotient Quotient s₂, provided hp₂ and h witness the needed compatibilities.
Русский
Пусть p₁ — предикат на α, s₁ — множество эквивалентности на α, s₂ — множество эквивалентности на подпредикатe p₁, p₂ — предикат на Quotient s₁, и dаны совместимости hp₂ и h. Тогда существует эквива́нтность между подтипом {x ∈ α | p₂ x} и частным quotient Quotient s₂, при условии hp₂ и h.
LaTeX
$$$\\left\\{ x \\in α \\mid p₂(x) \\right\\} \\simeq \\operatorname{Quotient}(s₂)$, \\\\ \\\\ \text{defined by } \\text{(forward/backward mappings as in subtypeQuotientEquivQuotientSubtype with hp₂, h)}$$
Lean4
theorem subtypeEquivCodomain_symm_apply_eq (f : { x' // x' ≠ x } → Y) (y : Y) :
((subtypeEquivCodomain f).symm y : X → Y) x = y :=
dif_neg (not_not.mpr rfl)
