83 lines
3 KiB
Text
83 lines
3 KiB
Text
/-
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Copyright (c) 2015 Microsoft Corporation. All rights reserved.
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Released under Apache 2.0 license as described in the file LICENSE.
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Module: data.examples.depchoice
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Author: Leonardo de Moura
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-/
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import data.encodable
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open nat encodable
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/-
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In mathematics, the axiom of dependent choice is a weak form of the axiom of choice that is
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sufficient to develop most of real analysis. See http://en.wikipedia.org/wiki/Axiom_of_dependent_choice.
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We can state it as follows:
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-/
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definition dependent_choice {A : Type} (R : A → A → Prop) :=
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(∀ a : A, ∃ b : A, R a b) → (∀ a : A, ∃ f : nat → A, f 0 = a ∧ ∀ n, R (f n) (f (n+1)))
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/-
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If A is an encodable type, and R is a decidable relation, we can prove (dependent_choice R) using the
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constructive choice function "choose"
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-/
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section depchoice
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parameters {A : Type} {R : A → A → Prop}
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parameters [encA : encodable A] [decR : decidable_rel R]
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include encA decR
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local infix `~` := R
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private definition f_aux (a : A) (H : ∀ a, ∃ b, a ~ b) : nat → A
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| 0 := a
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| (n+1) := choose (H (f_aux n))
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theorem dependent_choice_of_encodable_of_decidable : dependent_choice R :=
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assume H : ∀ a, ∃ b, a ~ b,
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take a : A,
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let f : nat → A := f_aux a H in
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have f_zero : f 0 = a, from rfl,
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have R_seq : ∀ n, f n ~ f (n+1), from
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take n, show f n ~ choose (H (f n)), from !choose_spec,
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exists.intro f (and.intro f_zero R_seq)
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/-
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The following slightly stronger version can be proved, where we also "return" the constructed function f.
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We just have to use Σ instead of ∃, and use Σ-constructor instead of exists.intro.
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Recall that ⟨f, H⟩ is notation for (sigma.mk f H)
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-/
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theorem stronger_dependent_choice_of_encodable_of_decidable
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: (∀ a, ∃ b, R a b) → (∀ a, Σ f, f 0 = a ∧ ∀ n, f n ~ f (n+1)) :=
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assume H : ∀ a, ∃ b, a ~ b,
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take a : A,
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let f : nat → A := f_aux a H in
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have f_zero : f 0 = a, from rfl,
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have R_seq : ∀ n, f n ~ f (n+1), from
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take n, show f n ~ choose (H (f n)), from !choose_spec,
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⟨f, and.intro f_zero R_seq⟩
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end depchoice
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/-
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If we encode dependent_choice using Σ instead of ∃.
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Then, we can prove this version without using any extra hypothesis (e.g., A is encodable or R is decidable).
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The function f can be constructed directly from the hypothesis: ∀ a : A, Σ b : A, R a b
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because Σ "carries" the witness 'b'. That is, we don't have to search for anything using "choose".
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-/
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open sigma.ops
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section sigma_depchoice
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parameters {A : Type} {R : A → A → Prop}
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local infix `~` := R
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private definition f_aux (a : A) (H : ∀ a, Σ b, a ~ b) : nat → A
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| 0 := a
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| (n+1) := (H (f_aux n)).1
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theorem sigma_dependent_choice : (∀ a, Σ b, R a b) → (∀ a, Σ f, f 0 = a ∧ ∀ n, f n ~ f (n+1)) :=
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assume H : ∀ a, Σ b, a ~ b,
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take a : A,
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let f : nat → A := f_aux a H in
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have f_zero : f 0 = a, from rfl,
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have R_seq : ∀ n, f n ~ f (n+1), from take n, (H (f n)).2,
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⟨f, and.intro f_zero R_seq⟩
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end sigma_depchoice
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