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Added a discussion of set-theoretic models of subtyping

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ajeffrey@roblox.com 2022-03-25 13:05:34 -05:00
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commit ec7c5e3b84
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prototyping/Properties/Subtyping.agda
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@ -9,6 +9,7 @@ open import Luau.Type using (Type; Scalar; strict; nil; number; string; boolean;
open import Properties.Contradiction using (CONTRADICTION; ¬)
open import Properties.Equality using (_≢_)
open import Properties.Functions using (_∘_)
open import Properties.Product using (_×_; _,_)
src = Luau.Type.src strict
@ -219,3 +220,65 @@ any-≮:-none = witness (scalar nil) any none
function-≮:-none : ∀ {T U} → ((T ⇒ U) ≮: none)
function-≮:-none = witness function function none
-- A Gentle Introduction To Semantic Subtyping (https://www.cduce.org/papers/gentle.pdf)
-- defines a "set-theoretic" model (sec 2.5)
-- Unfortunately we don't quite have this property, due to uninhabited types,
-- for example (none -> T) is equivalent to (none -> U)
-- when types are interpreted as sets of syntactic values.
_⊆_ : ∀ {A : Set} → (A → Set) → (A → Set) → Set
(P ⊆ Q) = ∀ a → (P a) → (Q a)
_⊗_ : ∀ {A B : Set} → (A → Set) → (B → Set) → ((A × B) → Set)
(P ⊗ Q) (a , b) = (P a) × (Q b)
Comp : ∀ {A : Set} → (A → Set) → (A → Set)
Comp P a = ¬(P a)
set-theoretic-if : ∀ {S₁ T₁ S₂ T₂} →
-- This is the "if" part of being a set-theoretic model
(Language (S₁ ⇒ T₁) ⊆ Language (S₂ ⇒ T₂)) →
(∀ Q → Q ⊆ Comp((Language S₁) ⊗ Comp(Language T₁)) → Q ⊆ Comp((Language S₂) ⊗ Comp(Language T₂)))
set-theoretic-if {S₁} {T₁} {S₂} {T₂} p Q q (t , u) Qtu (S₂t , ¬T₂u) = q (t , u) Qtu (S₁t , ¬T₁u) where
S₁t : Language S₁ t
S₁t with dec-language S₁ t
S₁t | Left ¬S₁t with p (function-err t) (function-err ¬S₁t)
S₁t | Left ¬S₁t | function-err ¬S₂t = CONTRADICTION (language-comp t ¬S₂t S₂t)
S₁t | Right r = r
¬T₁u : ¬(Language T₁ u)
¬T₁u T₁u with p (function-ok u) (function-ok T₁u)
¬T₁u T₁u | function-ok T₂u = ¬T₂u T₂u
not-quite-set-theoretic-only-if : ∀ {S₁ T₁ S₂ T₂} →
-- We don't quite have that this is a set-theoretic model
-- it's only true when Language T₁ and ¬Language T₂ t₂ are inhabited
-- in particular it's not true when T₁ is none, or T₂ is any.
∀ s₂ t₂ → Language S₂ s₂ → ¬Language T₂ t₂ →
-- This is the "only if" part of being a set-theoretic model
(∀ Q → Q ⊆ Comp((Language S₁) ⊗ Comp(Language T₁)) → Q ⊆ Comp((Language S₂) ⊗ Comp(Language T₂))) →
(Language (S₁ ⇒ T₁) ⊆ Language (S₂ ⇒ T₂))
not-quite-set-theoretic-only-if {S₁} {T₁} {S₂} {T₂} s₂ t₂ S₂s₂ ¬T₂t₂ p = {!!} where
Q : (Tree × Tree) → Set
Q (t , u) = Either (¬Language S₁ t) (Language T₁ u)
q : Q ⊆ Comp((Language S₁) ⊗ Comp(Language T₁))
q (t , u) (Left ¬S₁t) (S₁t , ¬T₁u) = language-comp t ¬S₁t S₁t
q (t , u) (Right T₂u) (S₁t , ¬T₁u) = ¬T₁u T₂u
r : Language (S₁ ⇒ T₁) ⊆ Language (S₂ ⇒ T₂)
r function function = function
r (function-err t) (function-err ¬S₁t) with dec-language S₂ t
r (function-err t) (function-err ¬S₁t) | Left ¬S₂t = function-err ¬S₂t
r (function-err t) (function-err ¬S₁t) | Right S₂t = CONTRADICTION (p Q q (t , t₂) (Left ¬S₁t) (S₂t , language-comp t₂ ¬T₂t₂))
r (function-ok t) (function-ok T₁t) with dec-language T₂ t
r (function-ok t) (function-ok T₁t) | Left ¬T₂t = CONTRADICTION (p Q q (s₂ , t) (Right T₁t) (S₂s₂ , language-comp t ¬T₂t))
r (function-ok t) (function-ok T₁t) | Right T₂t = function-ok T₂t