test(tests/lean/run): pre-quotient experiment

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Leonardo de Moura 2014-11-17 18:30:11 -08:00
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tests/lean/run/pquot.lean Normal file
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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 (heq.to_eq 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.from_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)