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There is exactly one way to choose 0 elements from the empty set.
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module Choose where | |
open import Data.Product using (_,_ ; proj₁ ; proj₂ ; ∃) renaming (_×_ to _∧_) | |
open import Relation.Binary.PropositionalEquality using (_≡_ ; sym ; subst) | |
open import Relation.Nullary using (¬_) | |
infix 0 _⇔_ | |
_⇔_ : Set → Set → Set | |
A ⇔ B = (A → B) ∧ (B → A) | |
postulate | |
set : Set | |
_∈_ : set → set → Set | |
_⊆_ : set → set → Set | |
a ⊆ b = ∀ x → x ∈ a → x ∈ b | |
postulate | |
ZF-extensionality : ∀ {x y} → (∀ z → z ∈ x ⇔ z ∈ y) → x ≡ y | |
ZF-specification : ∀ (P : set → Set) z → ∃ (λ y → ∀ x → x ∈ y ⇔ x ∈ z ∧ P x) | |
ZF-pairing : ∀ x y → ∃ (λ z → x ∈ z ∧ y ∈ z) | |
ZF-powerset : ∀ x → ∃ (λ y → ∀ z → z ⊆ x → z ∈ y) | |
ZF-emptyset : ∃ (λ x → ∀ y → ¬ y ∈ x) | |
proper-powerset : ∀ x → ∃ (λ y → ∀ z → z ⊆ x ⇔ z ∈ y) | |
proper-powerset x = | |
let px , f = ZF-powerset x in | |
let ppx , g = ZF-specification (_⊆ x) px in | |
ppx , λ y → | |
(λ y⊆x → proj₂ (g y) (f y y⊆x , y⊆x)) , | |
(λ y∈ppx → proj₂ (proj₁ (g y) y∈ppx)) | |
subsets-of : set → set | |
subsets-of x = proj₁ (proper-powerset x) | |
specified : (set → Set) → set → set | |
specified P x = proj₁ (ZF-specification P x) | |
Empty : set | |
Empty = proj₁ ZF-emptyset | |
Zero : set | |
Zero = Empty | |
one : ∃ (λ One → ∀ z → z ∈ One ⇔ z ≡ Zero) | |
one = | |
let pzz , a , b = ZF-pairing Zero Zero in | |
let One , g = ZF-specification (_≡ Zero) pzz in | |
One , λ z → | |
(λ z∈1 → proj₂ (proj₁ (g z) z∈1)) , | |
(λ z≡0 → proj₂ (g z) (subst (_∈ _) (sym z≡0) a , z≡0)) | |
One : set | |
One = proj₁ one | |
lem₁ : ∀ {z} → z ≡ Empty → z ⊆ Empty | |
lem₁ z≡0 x x∈z = subst (x ∈_) z≡0 x∈z | |
lem₂ : ∀ {z} → z ≡ Empty → z ∈ subsets-of Empty | |
lem₂ {z} z≡0 = proj₁ (proj₂ (proper-powerset Empty) z) (lem₁ z≡0) | |
thm : specified (_≡ Zero) (subsets-of Empty) ≡ One | |
thm = | |
let h1 = proj₂ (ZF-specification (_≡ Zero) (subsets-of Empty)) in | |
ZF-extensionality λ z → | |
(λ z∈s → | |
let z≡0 = proj₂ (proj₁ (h1 z) z∈s) in | |
proj₂ (proj₂ one z) z≡0) , | |
(λ z∈1 → | |
let z≡0 = proj₁ (proj₂ one z) z∈1 in | |
proj₂ (h1 z) (lem₂ z≡0 , z≡0)) |
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