import logic.cast data.list data.sigma -- The (pre-)quotient type kernel extension would add the following constants -- quot, pquot.mk, pquot.eqv and pquot.rec -- and a computational rule, which we call pquot.comp here. -- Note that, these constants do not assume the environment contains = constant pquot.{l} {A : Type.{l}} (R : A → A → Prop) : Type.{l} constant pquot.abs {A : Type} (R : A → A → Prop) : A → pquot R -- pquot.eqv is a way to say R a b → (pquot.abs R a) = (pquot.abs R b) without mentioning equality constant pquot.eqv {A : Type} (R : A → A → Prop) {a b : A} : R a b → ∀ (P : pquot R → Prop), P (pquot.abs R a) → P (pquot.abs R b) constant pquot.rec {A : Type} {R : A → A → Prop} {C : pquot R → Type} (f : Π a, C (pquot.abs R a)) -- sound is essentially saying: ∀ (a b : A) (H : R a b), f a == f b -- H makes sure we can only define a function on (quot R) if for all a b : A -- R a b → f a == f b (sound : ∀ a b, R a b → ∀ P : (Π (q : pquot R), C q → Prop), P (pquot.abs R a) (f a) → P (pquot.abs R b) (f b)) (q : pquot R) : C q -- We would also get the following computational rule: -- pquot.rec R H₁ H₂ (pquot.abs R a) ==> H₁ a constant pquot.comp {A : Type} {R : A → A → Prop} {C : pquot R → Type} (f : Π a, C (pquot.abs R a)) (sound : ∀ a b, R a b → ∀ P : (Π (q : pquot R), C q → Prop), P (pquot.abs R a) (f a) → P (pquot.abs R b) (f b)) (a : A) -- In the implementation this would be a computational rule : pquot.rec f sound (pquot.abs R a) = f a -- If the environment contains = and ==, then we can define definition pquot.eq {A : Type} (R : A → A → Prop) {a b : A} (H : R a b) : pquot.abs R a = pquot.abs R b := have aux : ∀ (P : pquot R → Prop), P (pquot.abs R a) → P (pquot.abs R b), from pquot.eqv R H, aux (λ x : pquot R, pquot.abs R a = x) rfl definition pquot.rec_on {A : Type} {R : A → A → Prop} {C : pquot R → Type} (q : pquot R) (f : Π a, C (pquot.abs R a)) (sound : ∀ (a b : A), R a b → f a == f b) : C q := pquot.rec f (λ (a b : A) (H : R a b) (P : Π (q : pquot R), C q → Prop) (Ha : P (pquot.abs R a) (f a)), have aux₁ : f a == f b, from sound a b H, have aux₂ : pquot.abs R a = pquot.abs R b, from pquot.eq R H, have aux₃ : ∀ (c₁ c₂ : C (pquot.abs R a)) (e : c₁ == c₂), P (pquot.abs R a) c₁ → P (pquot.abs R a) c₂, from λ c₁ c₂ e H, eq.rec_on (eq_of_heq e) H, have aux₄ : ∀ (c₁ : C (pquot.abs R a)) (c₂ : C (pquot.abs R b)) (e : c₁ == c₂), P (pquot.abs R a) c₁ → P (pquot.abs R b) c₂, from eq.rec_on aux₂ aux₃, show P (pquot.abs R b) (f b), from aux₄ (f a) (f b) aux₁ Ha) q definition pquot.lift {A : Type} {R : A → A → Prop} {B : Type} (f : A → B) (sound : ∀ (a b : A), R a b → f a = f b) (q : pquot R) : B := pquot.rec_on q f (λ (a b : A) (H : R a b), heq_of_eq (sound a b H)) theorem pquot.induction_on {A : Type} {R : A → A → Prop} {P : pquot R → Prop} (q : pquot R) (f : ∀ a, P (pquot.abs R a)) : P q := pquot.rec_on q f (λ (a b : A) (H : R a b), have aux₁ : pquot.abs R a = pquot.abs R b, from pquot.eq R H, have aux₂ : P (pquot.abs R a) = P (pquot.abs R b), from congr_arg P aux₁, have aux₃ : cast aux₂ (f a) = f b, from proof_irrel (cast aux₂ (f a)) (f b), show f a == f b, from @cast_to_heq _ _ _ _ aux₂ aux₃) theorem pquot.abs.surjective {A : Type} {R : A → A → Prop} : ∀ q : pquot R, ∃ x : A, pquot.abs R x = q := take q, pquot.induction_on q (take a, exists.intro a rfl) definition pquot.exact {A : Type} (R : A → A → Prop) := ∀ a b : A, pquot.abs R a = pquot.abs R b → R a b -- Definable quotient structure dquot {A : Type} (R : A → A → Prop) := mk :: (rep : pquot R → A) (complete : ∀a, R (rep (pquot.abs R a)) a) -- (stable : ∀q, pquot.abs R (rep q) = q) structure is_equiv {A : Type} (R : A → A → Prop) := mk :: (refl : ∀x, R x x) (symm : ∀{x y}, R x y → R y x) (trans : ∀{x y z}, R x y → R y z → R x z) -- Definiable quotients are exact if R is an equivalence relation theorem quot_exact {A : Type} {R : A → A → Prop} (eqv : is_equiv R) (q : dquot R) : pquot.exact R := λ (a b : A) (H : pquot.abs R a = pquot.abs R b), have H₁ : pquot.abs R a = pquot.abs R a → R (dquot.rep q (pquot.abs R a)) (dquot.rep q (pquot.abs R a)), from λH, is_equiv.refl eqv _, have H₂ : pquot.abs R a = pquot.abs R b → R (dquot.rep q (pquot.abs R a)) (dquot.rep q (pquot.abs R b)), from eq.subst H H₁, have H₃ : R (dquot.rep q (pquot.abs R a)) (dquot.rep q (pquot.abs R b)), from H₂ H, have H₄ : R a (dquot.rep q (pquot.abs R a)), from is_equiv.symm eqv (dquot.complete q a), have H₅ : R (dquot.rep q (pquot.abs R b)) b, from dquot.complete q b, is_equiv.trans eqv H₄ (is_equiv.trans eqv H₃ H₅)