Prototype: Added a discussion of set-theoretic models of subtyping (#431)

* Added a discussion of set-theoretic models of subtyping to the prototype
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Alan Jeffrey 2022-03-31 18:29:42 -05:00 committed by GitHub
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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 +220,92 @@ 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 = r 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
-- A counterexample when the argument type is empty.
set-theoretic-counterexample-one : ( Q Q Comp((Language none) Comp(Language number)) Q Comp((Language none) Comp(Language string)))
set-theoretic-counterexample-one Q q ((scalar s) , u) Qtu (scalar () , p)
set-theoretic-counterexample-one Q q ((function-err t) , u) Qtu (scalar-function-err () , p)
set-theoretic-counterexample-two : (none number) ≮: (none string)
set-theoretic-counterexample-two = witness
(function-ok (scalar number)) (function-ok (scalar number))
(function-ok (scalar-scalar number string (λ ())))
-- At some point we may deal with overloaded function resolution, which should fix this problem...
-- The reason why this is connected to overloaded functions is that currently we have that the type of
-- f(x) is (tgt T) where f:T. Really we should have the type depend on the type of x, that is use (tgt T U),
-- where U is the type of x. In particular (tgt (S => T) (U & V)) should be the same as (tgt ((S&U) => T) V)
-- and tgt(none => T) should be any. For example
--
-- tgt((number => string) & (string => bool))(number)
-- is tgt(number => string)(number) & tgt(string => bool)(number)
-- is tgt(number => string)(number) & tgt(string => bool)(number&any)
-- is tgt(number => string)(number) & tgt(string&number => bool)(any)
-- is tgt(number => string)(number) & tgt(none => bool)(any)
-- is string & any
-- is string
--
-- there's some discussion of this in the Gentle Introduction paper.