SubSubtyping
Set Warnings "-notation-overridden,-parsing,-deprecated-hint-without-locality".
From Coq Require Import Strings.String.
From PLF Require Import Maps.
From PLF Require Import Types.
From PLF Require Import Smallstep.
From Coq Require Import Strings.String.
From PLF Require Import Maps.
From PLF Require Import Types.
From PLF Require Import Smallstep.
Concepts
A Motivating Example
Person = {name:String, age:Nat} Student = {name:String, age:Nat, gpa:Nat}
(\r:Person. (r.age)+1) {name="Pat",age=21,gpa=1}is not typable, since it applies a function that wants a two-field record to an argument that actually provides three fields, while the T_App rule demands that the domain type of the function being applied must match the type of the argument precisely.
- S is a subtype of T, written S <: T, if a value of type S can safely be used in any context where a value of type T is expected.
Subtyping and Object-Oriented Languages
The Subsumption Rule
- Defining a binary subtype relation between types.
- Enriching the typing relation to take subtyping into account.
Gamma ⊢ t_{1} ∈ T_{1} T_{1} <: T_{2} | (T_Sub) |
Gamma ⊢ t_{1} ∈ T_{2} |
The Subtype Relation
Structural Rules
S <: U U <: T | (S_Trans) |
S <: T |
(S_Refl) | |
T <: T |
Products
S_{1} <: T_{1} S_{2} <: T_{2} | (S_Prod) |
S_{1} * S_{2} <: T_{1} * T_{2} |
Arrows
f : C → Student
g : (C→Person) → D That is, f is a function that yields a record of type Student, and g is a (higher-order) function that expects its argument to be a function yielding a record of type Person. Also suppose that Student is a subtype of Person. Then the application g f is safe even though their types do not match up precisely, because the only thing g can do with f is to apply it to some argument (of type C); the result will actually be a Student, while g will be expecting a Person, but this is safe because the only thing g can then do is to project out the two fields that it knows about (name and age), and these will certainly be among the fields that are present.
S_{2} <: T_{2} | (S_Arrow_Co) |
S_{1} -> S_{2} <: S_{1} -> T_{2} |
T_{1} <: S_{1} S_{2} <: T_{2} | (S_Arrow) |
S_{1} -> S_{2} <: T_{1} -> T_{2} |
f : Person → C
g : (Student → C) → D The application g f is safe, because the only thing the body of g can do with f is to apply it to some argument of type Student. Since f requires records having (at least) the fields of a Person, this will always work. So Person → C is a subtype of Student → C since Student is a subtype of Person.
Records
{name:String, age:Nat, gpa:Nat} <: {name:String, age:Nat}
{name:String, age:Nat} <: {name:String}
{name:String} <: {} This is known as "width subtyping" for records.
{x:Student} <: {x:Person} This is known as "depth subtyping".
{name:String,age:Nat} <: {age:Nat,name:String} This is known as "permutation subtyping".
forall jk in j_{1}..jn, | |
exists ip in i_{1}..im, such that | |
jk=ip and Sp <: Tk | (S_Rcd) |
{i_{1}:S_{1}...im:Sm} <: {j_{1}:T_{1}...jn:Tn} |
n > m | (S_RcdWidth) |
{i_{1}:T_{1}...in:Tn} <: {i_{1}:T_{1}...im:Tm} |
S_{1} <: T_{1} ... Sn <: Tn | (S_RcdDepth) |
{i_{1}:S_{1}...in:Sn} <: {i_{1}:T_{1}...in:Tn} |
{i_{1}:S_{1}...in:Sn} is a permutation of {j_{1}:T_{1}...jn:Tn} | (S_RcdPerm) |
{i_{1}:S_{1}...in:Sn} <: {j_{1}:T_{1}...jn:Tn} |
- Each class member (field or method) can be assigned a single
index, adding new indices "on the right" as more members are
added in subclasses (i.e., no permutation for classes).
- A class may implement multiple interfaces -- so-called "multiple
inheritance" of interfaces (i.e., permutation is allowed for
interfaces).
- In early versions of Java, a subclass could not change the argument or result types of a method of its superclass (i.e., no depth subtyping or no arrow subtyping, depending how you look at it).
Exercise: 2 stars, standard, especially useful (arrow_sub_wrong)
Suppose we had incorrectly defined subtyping as covariant on both the right and the left of arrow types:S_{1} <: T_{1} S_{2} <: T_{2} | (S_Arrow_wrong) |
S_{1} -> S_{2} <: T_{1} -> T_{2} |
f : Student → Nat
g : (Person → Nat) → Nat ... such that the application g f will get stuck during execution. (Use informal syntax. No need to prove formally that the application gets stuck.)
(* Do not modify the following line: *)
Definition manual_grade_for_arrow_sub_wrong : option (nat×string) := None.
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Definition manual_grade_for_arrow_sub_wrong : option (nat×string) := None.
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Top
(S_Top) | |
S <: Top |
Summary
- adding a base type Top,
- adding the rule of subsumption
to the typing relation, andGamma ⊢ t_{1} ∈ T_{1} T_{1} <: T_{2} (T_Sub) Gamma ⊢ t_{1} ∈ T_{2} - defining a subtype relation as follows:
S <: U U <: T (S_Trans) S <: T (S_Refl) T <: T (S_Top) S <: Top S_{1} <: T_{1} S_{2} <: T_{2} (S_Prod) S_{1} * S_{2} <: T_{1} * T_{2} T_{1} <: S_{1} S_{2} <: T_{2} (S_Arrow) S_{1} -> S_{2} <: T_{1} -> T_{2} n > m (S_RcdWidth) {i_{1}:T_{1}...in:Tn} <: {i_{1}:T_{1}...im:Tm} S_{1} <: T_{1} ... Sn <: Tn (S_RcdDepth) {i_{1}:S_{1}...in:Sn} <: {i_{1}:T_{1}...in:Tn} {i_{1}:S_{1}...in:Sn} is a permutation of {j_{1}:T_{1}...jn:Tn} (S_RcdPerm) {i_{1}:S_{1}...in:Sn} <: {j_{1}:T_{1}...jn:Tn}
Exercises
Exercise: 1 star, standard, optional (subtype_instances_tf_1)
Suppose we have types S, T, U, and V with S <: T and U <: V. Which of the following subtyping assertions are then true? Write true or false after each one. (A, B, and C here are base types like Bool, Nat, etc.)- T→S <: T→S
- Top→U <: S→Top
- (C→C) → (A×B) <: (C→C) → (Top×B)
- T→T→U <: S→S→V
- (T→T)->U <: (S→S)->V
- ((T→S)->T)->U <: ((S→T)->S)->V
- S×V <: T×U
Exercise: 2 stars, standard (subtype_order)
The following types happen to form a linear order with respect to subtyping:- Top
- Top → Student
- Student → Person
- Student → Top
- Person → Student
(* Do not modify the following line: *)
Definition manual_grade_for_subtype_order : option (nat×string) := None.
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Definition manual_grade_for_subtype_order : option (nat×string) := None.
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Exercise: 1 star, standard (subtype_instances_tf_2)
Which of the following statements are true? Write true or false after each one.∀ S T,
S <: T →
S→S <: T→T
∀ S,
S <: A→A →
∃ T,
S = T→T ∧ T <: A
∀ S T_{1} T_{2},
(S <: T_{1} → T_{2}) →
∃ S_{1} S_{2},
S = S_{1} → S_{2} ∧ T_{1} <: S_{1} ∧ S_{2} <: T_{2}
∃ S,
S <: S→S
∃ S,
S→S <: S
∀ S T_{1} T_{2},
S <: T_{1}×T_{2} →
∃ S_{1} S_{2},
S = S_{1}×S_{2} ∧ S_{1} <: T_{1} ∧ S_{2} <: T_{2}
(* Do not modify the following line: *)
Definition manual_grade_for_subtype_instances_tf_2 : option (nat×string) := None.
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Definition manual_grade_for_subtype_instances_tf_2 : option (nat×string) := None.
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Exercise: 1 star, standard (subtype_concepts_tf)
Which of the following statements are true, and which are false?- There exists a type that is a supertype of every other type.
- There exists a type that is a subtype of every other type.
- There exists a pair type that is a supertype of every other
pair type.
- There exists a pair type that is a subtype of every other
pair type.
- There exists an arrow type that is a supertype of every other
arrow type.
- There exists an arrow type that is a subtype of every other
arrow type.
- There is an infinite descending chain of distinct types in the
subtype relation---that is, an infinite sequence of types
S_{0}, S_{1}, etc., such that all the Si's are different and
each S(i+1) is a subtype of Si.
- There is an infinite ascending chain of distinct types in the subtype relation---that is, an infinite sequence of types S_{0}, S_{1}, etc., such that all the Si's are different and each S(i+1) is a supertype of Si.
(* Do not modify the following line: *)
Definition manual_grade_for_subtype_concepts_tf : option (nat×string) := None.
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Definition manual_grade_for_subtype_concepts_tf : option (nat×string) := None.
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Exercise: 2 stars, standard (proper_subtypes)
Is the following statement true or false? Briefly explain your answer. (Here Base n stands for a base type, where n is a string standing for the name of the base type. See the Syntax section below.)∀ T,
~(T = Bool ∨ ∃ n, T = Base n) →
∃ S,
S <: T ∧ S ≠ T
(* Do not modify the following line: *)
Definition manual_grade_for_proper_subtypes : option (nat×string) := None.
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Definition manual_grade_for_proper_subtypes : option (nat×string) := None.
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Exercise: 2 stars, standard (small_large_1)
- What is the smallest type T ("smallest" in the subtype
relation) that makes the following assertion true? (Assume we
have Unit among the base types and unit as a constant of this
type.)
empty ⊢ (\p:T×Top. p.fst) ((\z:A.z), unit) \in A→A - What is the largest type T that makes the same assertion true?
(* Do not modify the following line: *)
Definition manual_grade_for_small_large_1 : option (nat×string) := None.
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Definition manual_grade_for_small_large_1 : option (nat×string) := None.
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Exercise: 2 stars, standard (small_large_2)
- What is the smallest type T that makes the following
assertion true?
empty ⊢ (\p:(A→A × B→B). p) ((\z:A.z), (\z:B.z)) \in T - What is the largest type T that makes the same assertion true?
(* Do not modify the following line: *)
Definition manual_grade_for_small_large_2 : option (nat×string) := None.
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Definition manual_grade_for_small_large_2 : option (nat×string) := None.
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Exercise: 2 stars, standard, optional (small_large_3)
- What is the smallest type T that makes the following
assertion true?
a:A ⊢ (\p:(A×T). (p.snd) (p.fst)) (a, \z:A.z) \in A - What is the largest type T that makes the same assertion true?
Exercise: 2 stars, standard (small_large_4)
- What is the smallest type T (if one exists) that makes the
following assertion true?
∃ S,
empty ⊢ (\p:(A×T). (p.snd) (p.fst)) \in S - What is the largest type T that makes the same assertion true?
(* Do not modify the following line: *)
Definition manual_grade_for_small_large_4 : option (nat×string) := None.
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Definition manual_grade_for_small_large_4 : option (nat×string) := None.
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Exercise: 2 stars, standard (smallest_1)
What is the smallest type T (if one exists) that makes the following assertion true?∃ S t,
empty ⊢ (\x:T. x x) t \in S
(* Do not modify the following line: *)
Definition manual_grade_for_smallest_1 : option (nat×string) := None.
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Definition manual_grade_for_smallest_1 : option (nat×string) := None.
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Exercise: 2 stars, standard (smallest_2)
What is the smallest type T that makes the following assertion true?empty ⊢ (\x:Top. x) ((\z:A.z) , (\z:B.z)) \in T
(* Do not modify the following line: *)
Definition manual_grade_for_smallest_2 : option (nat×string) := None.
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Definition manual_grade_for_smallest_2 : option (nat×string) := None.
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Exercise: 3 stars, standard, optional (count_supertypes)
How many supertypes does the record type {x:A, y:C→C} have? That is, how many different types T are there such that {x:A, y:C→C} <: T? (We consider two types to be different if they are written differently, even if each is a subtype of the other. For example, {x:A,y:B} and {y:B,x:A} are different.)Exercise: 2 stars, standard (pair_permutation)
The subtyping rule for product typesS_{1} <: T_{1} S_{2} <: T_{2} | (S_Prod) |
S_{1}*S_{2} <: T_{1}*T_{2} |
T_{1}*T_{2} <: T_{2}*T_{1} |
(* Do not modify the following line: *)
Definition manual_grade_for_pair_permutation : option (nat×string) := None.
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Definition manual_grade_for_pair_permutation : option (nat×string) := None.
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Formal Definitions
Syntax
Inductive ty : Type :=
| Ty_Top : ty
| Ty_Bool : ty
| Ty_Base : string → ty
| Ty_Arrow : ty → ty → ty
| Ty_Unit : ty
.
Inductive tm : Type :=
| tm_var : string → tm
| tm_app : tm → tm → tm
| tm_abs : string → ty → tm → tm
| tm_true : tm
| tm_false : tm
| tm_if : tm → tm → tm → tm
| tm_unit : tm
.
Declare Custom Entry stlc.
Notation "<{ e }>" := e (e custom stlc at level 99).
Notation "( x )" := x (in custom stlc, x at level 99).
Notation "x" := x (in custom stlc at level 0, x constr at level 0).
Notation "S -> T" := (Ty_Arrow S T) (in custom stlc at level 50, right associativity).
Notation "x y" := (tm_app x y) (in custom stlc at level 1, left associativity).
Notation "\ x : t , y" :=
(tm_abs x t y) (in custom stlc at level 90, x at level 99,
t custom stlc at level 99,
y custom stlc at level 99,
left associativity).
Coercion tm_var : string >-> tm.
Notation "'Bool'" := Ty_Bool (in custom stlc at level 0).
Notation "'if' x 'then' y 'else' z" :=
(tm_if x y z) (in custom stlc at level 89,
x custom stlc at level 99,
y custom stlc at level 99,
z custom stlc at level 99,
left associativity).
Notation "'true'" := true (at level 1).
Notation "'true'" := tm_true (in custom stlc at level 0).
Notation "'false'" := false (at level 1).
Notation "'false'" := tm_false (in custom stlc at level 0).
Notation "'Unit'" :=
(Ty_Unit) (in custom stlc at level 0).
Notation "'unit'" := tm_unit (in custom stlc at level 0).
Notation "'Base' x" := (Ty_Base x) (in custom stlc at level 0).
Notation "'Top'" := (Ty_Top) (in custom stlc at level 0).
| Ty_Top : ty
| Ty_Bool : ty
| Ty_Base : string → ty
| Ty_Arrow : ty → ty → ty
| Ty_Unit : ty
.
Inductive tm : Type :=
| tm_var : string → tm
| tm_app : tm → tm → tm
| tm_abs : string → ty → tm → tm
| tm_true : tm
| tm_false : tm
| tm_if : tm → tm → tm → tm
| tm_unit : tm
.
Declare Custom Entry stlc.
Notation "<{ e }>" := e (e custom stlc at level 99).
Notation "( x )" := x (in custom stlc, x at level 99).
Notation "x" := x (in custom stlc at level 0, x constr at level 0).
Notation "S -> T" := (Ty_Arrow S T) (in custom stlc at level 50, right associativity).
Notation "x y" := (tm_app x y) (in custom stlc at level 1, left associativity).
Notation "\ x : t , y" :=
(tm_abs x t y) (in custom stlc at level 90, x at level 99,
t custom stlc at level 99,
y custom stlc at level 99,
left associativity).
Coercion tm_var : string >-> tm.
Notation "'Bool'" := Ty_Bool (in custom stlc at level 0).
Notation "'if' x 'then' y 'else' z" :=
(tm_if x y z) (in custom stlc at level 89,
x custom stlc at level 99,
y custom stlc at level 99,
z custom stlc at level 99,
left associativity).
Notation "'true'" := true (at level 1).
Notation "'true'" := tm_true (in custom stlc at level 0).
Notation "'false'" := false (at level 1).
Notation "'false'" := tm_false (in custom stlc at level 0).
Notation "'Unit'" :=
(Ty_Unit) (in custom stlc at level 0).
Notation "'unit'" := tm_unit (in custom stlc at level 0).
Notation "'Base' x" := (Ty_Base x) (in custom stlc at level 0).
Notation "'Top'" := (Ty_Top) (in custom stlc at level 0).
Reserved Notation "'[' x ':=' s ']' t" (in custom stlc at level 20, x constr).
Fixpoint subst (x : string) (s : tm) (t : tm) : tm :=
match t with
| tm_var y ⇒
if eqb_string x y then s else t
| <{\y:T, t_{1}}> ⇒
if eqb_string x y then t else <{\y:T, [x:=s] t_{1}}>
| <{t_{1} t_{2}}> ⇒
<{([x:=s] t_{1}) ([x:=s] t_{2})}>
| <{true}> ⇒
<{true}>
| <{false}> ⇒
<{false}>
| <{if t_{1} then t_{2} else t_{3}}> ⇒
<{if ([x:=s] t_{1}) then ([x:=s] t_{2}) else ([x:=s] t_{3})}>
| <{unit}> ⇒
<{unit}>
end
where "'[' x ':=' s ']' t" := (subst x s t) (in custom stlc).
Fixpoint subst (x : string) (s : tm) (t : tm) : tm :=
match t with
| tm_var y ⇒
if eqb_string x y then s else t
| <{\y:T, t_{1}}> ⇒
if eqb_string x y then t else <{\y:T, [x:=s] t_{1}}>
| <{t_{1} t_{2}}> ⇒
<{([x:=s] t_{1}) ([x:=s] t_{2})}>
| <{true}> ⇒
<{true}>
| <{false}> ⇒
<{false}>
| <{if t_{1} then t_{2} else t_{3}}> ⇒
<{if ([x:=s] t_{1}) then ([x:=s] t_{2}) else ([x:=s] t_{3})}>
| <{unit}> ⇒
<{unit}>
end
where "'[' x ':=' s ']' t" := (subst x s t) (in custom stlc).
Inductive value : tm → Prop :=
| v_abs : ∀ x T_{2} t_{1},
value <{\x:T_{2}, t_{1}}>
| v_true :
value <{true}>
| v_false :
value <{false}>
| v_unit :
value <{unit}>
.
Hint Constructors value : core.
Reserved Notation "t '-->' t'" (at level 40).
Inductive step : tm → tm → Prop :=
| ST_AppAbs : ∀ x T_{2} t_{1} v_{2},
value v_{2} →
<{(\x:T_{2}, t_{1}) v_{2}}> --> <{ [x:=v_{2}]t_{1} }>
| ST_App1 : ∀ t_{1} t_{1}' t_{2},
t_{1} --> t_{1}' →
<{t_{1} t_{2}}> --> <{t_{1}' t_{2}}>
| ST_App2 : ∀ v_{1} t_{2} t_{2}',
value v_{1} →
t_{2} --> t_{2}' →
<{v_{1} t_{2}}> --> <{v_{1} t_{2}'}>
| ST_IfTrue : ∀ t_{1} t_{2},
<{if true then t_{1} else t_{2}}> --> t_{1}
| ST_IfFalse : ∀ t_{1} t_{2},
<{if false then t_{1} else t_{2}}> --> t_{2}
| ST_If : ∀ t_{1} t_{1}' t_{2} t_{3},
t_{1} --> t_{1}' →
<{if t_{1} then t_{2} else t_{3}}> --> <{if t_{1}' then t_{2} else t_{3}}>
where "t '-->' t'" := (step t t').
Hint Constructors step : core.
| v_abs : ∀ x T_{2} t_{1},
value <{\x:T_{2}, t_{1}}>
| v_true :
value <{true}>
| v_false :
value <{false}>
| v_unit :
value <{unit}>
.
Hint Constructors value : core.
Reserved Notation "t '-->' t'" (at level 40).
Inductive step : tm → tm → Prop :=
| ST_AppAbs : ∀ x T_{2} t_{1} v_{2},
value v_{2} →
<{(\x:T_{2}, t_{1}) v_{2}}> --> <{ [x:=v_{2}]t_{1} }>
| ST_App1 : ∀ t_{1} t_{1}' t_{2},
t_{1} --> t_{1}' →
<{t_{1} t_{2}}> --> <{t_{1}' t_{2}}>
| ST_App2 : ∀ v_{1} t_{2} t_{2}',
value v_{1} →
t_{2} --> t_{2}' →
<{v_{1} t_{2}}> --> <{v_{1} t_{2}'}>
| ST_IfTrue : ∀ t_{1} t_{2},
<{if true then t_{1} else t_{2}}> --> t_{1}
| ST_IfFalse : ∀ t_{1} t_{2},
<{if false then t_{1} else t_{2}}> --> t_{2}
| ST_If : ∀ t_{1} t_{1}' t_{2} t_{3},
t_{1} --> t_{1}' →
<{if t_{1} then t_{2} else t_{3}}> --> <{if t_{1}' then t_{2} else t_{3}}>
where "t '-->' t'" := (step t t').
Hint Constructors step : core.
Subtyping
Reserved Notation "T '<:' U" (at level 40).
Inductive subtype : ty → ty → Prop :=
| S_Refl : ∀ T,
T <: T
| S_Trans : ∀ S U T,
S <: U →
U <: T →
S <: T
| S_Top : ∀ S,
S <: <{Top}>
| S_Arrow : ∀ S_{1} S_{2} T_{1} T_{2},
T_{1} <: S_{1} →
S_{2} <: T_{2} →
<{S_{1}→S_{2}}> <: <{T_{1}→T_{2}}>
where "T '<:' U" := (subtype T U).
Inductive subtype : ty → ty → Prop :=
| S_Refl : ∀ T,
T <: T
| S_Trans : ∀ S U T,
S <: U →
U <: T →
S <: T
| S_Top : ∀ S,
S <: <{Top}>
| S_Arrow : ∀ S_{1} S_{2} T_{1} T_{2},
T_{1} <: S_{1} →
S_{2} <: T_{2} →
<{S_{1}→S_{2}}> <: <{T_{1}→T_{2}}>
where "T '<:' U" := (subtype T U).
Note that we don't need any special rules for base types (Bool
and Base): they are automatically subtypes of themselves (by
S_Refl) and Top (by S_Top), and that's all we want.
Hint Constructors subtype : core.
Module Examples.
Open Scope string_scope.
Notation x := "x".
Notation y := "y".
Notation z := "z".
Notation A := <{Base "A"}>.
Notation B := <{Base "B"}>.
Notation C := <{Base "C"}>.
Notation String := <{Base "String"}>.
Notation Float := <{Base "Float"}>.
Notation Integer := <{Base "Integer"}>.
Example subtyping_example_0 :
<{C→Bool}> <: <{C→Top}>.
Proof. auto. Qed.
Module Examples.
Open Scope string_scope.
Notation x := "x".
Notation y := "y".
Notation z := "z".
Notation A := <{Base "A"}>.
Notation B := <{Base "B"}>.
Notation C := <{Base "C"}>.
Notation String := <{Base "String"}>.
Notation Float := <{Base "Float"}>.
Notation Integer := <{Base "Integer"}>.
Example subtyping_example_0 :
<{C→Bool}> <: <{C→Top}>.
Proof. auto. Qed.
Exercise: 2 stars, standard, optional (subtyping_judgements)
(Leave this exercise Admitted until after you have finished adding product types to the language -- see exercise products -- at least up to this point in the file).Person := { name : String }
Student := { name : String ; gpa : Float }
Employee := { name : String ; ssn : Integer }
Definition Person : ty
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Definition Student : ty
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Definition Employee : ty
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Definition Student : ty
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Definition Employee : ty
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Now use the definition of the subtype relation to prove the following:
Example sub_student_person :
Student <: Person.
Proof.
(* FILL IN HERE *) Admitted.
Example sub_employee_person :
Employee <: Person.
Proof.
(* FILL IN HERE *) Admitted.
☐
Student <: Person.
Proof.
(* FILL IN HERE *) Admitted.
Example sub_employee_person :
Employee <: Person.
Proof.
(* FILL IN HERE *) Admitted.
☐
Exercise: 1 star, standard, optional (subtyping_example_1)
Example subtyping_example_1 :
<{Top→Student}> <: <{(C→C)→Person}>.
Proof with eauto.
(* FILL IN HERE *) Admitted.
☐
<{Top→Student}> <: <{(C→C)→Person}>.
Proof with eauto.
(* FILL IN HERE *) Admitted.
☐
Example subtyping_example_2 :
<{Top→Person}> <: <{Person→Top}>.
Proof with eauto.
(* FILL IN HERE *) Admitted.
☐
<{Top→Person}> <: <{Person→Top}>.
Proof with eauto.
(* FILL IN HERE *) Admitted.
☐
Definition context := partial_map ty.
Reserved Notation "Gamma '⊢' t '∈' T" (at level 40,
t custom stlc, T custom stlc at level 0).
Inductive has_type : context → tm → ty → Prop :=
(* Same as before: *)
(* pure STLC *)
| T_Var : ∀ Gamma x T_{1},
Gamma x = Some T_{1} →
Gamma ⊢ x \in T_{1}
| T_Abs : ∀ Gamma x T_{1} T_{2} t_{1},
(x ⊢> T_{2} ; Gamma) ⊢ t_{1} \in T_{1} →
Gamma ⊢ \x:T_{2}, t_{1} \in (T_{2} → T_{1})
| T_App : ∀ T_{1} T_{2} Gamma t_{1} t_{2},
Gamma ⊢ t_{1} \in (T_{2} → T_{1}) →
Gamma ⊢ t_{2} \in T_{2} →
Gamma ⊢ t_{1} t_{2} \in T_{1}
| T_True : ∀ Gamma,
Gamma ⊢ true \in Bool
| T_False : ∀ Gamma,
Gamma ⊢ false \in Bool
| T_If : ∀ t_{1} t_{2} t_{3} T_{1} Gamma,
Gamma ⊢ t_{1} \in Bool →
Gamma ⊢ t_{2} \in T_{1} →
Gamma ⊢ t_{3} \in T_{1} →
Gamma ⊢ if t_{1} then t_{2} else t_{3} \in T_{1}
| T_Unit : ∀ Gamma,
Gamma ⊢ unit \in Unit
(* New rule of subsumption: *)
| T_Sub : ∀ Gamma t_{1} T_{1} T_{2},
Gamma ⊢ t_{1} \in T_{1} →
T_{1} <: T_{2} →
Gamma ⊢ t_{1} \in T_{2}
where "Gamma '⊢' t '∈' T" := (has_type Gamma t T).
Hint Constructors has_type : core.
Module Examples2.
Import Examples.
Reserved Notation "Gamma '⊢' t '∈' T" (at level 40,
t custom stlc, T custom stlc at level 0).
Inductive has_type : context → tm → ty → Prop :=
(* Same as before: *)
(* pure STLC *)
| T_Var : ∀ Gamma x T_{1},
Gamma x = Some T_{1} →
Gamma ⊢ x \in T_{1}
| T_Abs : ∀ Gamma x T_{1} T_{2} t_{1},
(x ⊢> T_{2} ; Gamma) ⊢ t_{1} \in T_{1} →
Gamma ⊢ \x:T_{2}, t_{1} \in (T_{2} → T_{1})
| T_App : ∀ T_{1} T_{2} Gamma t_{1} t_{2},
Gamma ⊢ t_{1} \in (T_{2} → T_{1}) →
Gamma ⊢ t_{2} \in T_{2} →
Gamma ⊢ t_{1} t_{2} \in T_{1}
| T_True : ∀ Gamma,
Gamma ⊢ true \in Bool
| T_False : ∀ Gamma,
Gamma ⊢ false \in Bool
| T_If : ∀ t_{1} t_{2} t_{3} T_{1} Gamma,
Gamma ⊢ t_{1} \in Bool →
Gamma ⊢ t_{2} \in T_{1} →
Gamma ⊢ t_{3} \in T_{1} →
Gamma ⊢ if t_{1} then t_{2} else t_{3} \in T_{1}
| T_Unit : ∀ Gamma,
Gamma ⊢ unit \in Unit
(* New rule of subsumption: *)
| T_Sub : ∀ Gamma t_{1} T_{1} T_{2},
Gamma ⊢ t_{1} \in T_{1} →
T_{1} <: T_{2} →
Gamma ⊢ t_{1} \in T_{2}
where "Gamma '⊢' t '∈' T" := (has_type Gamma t T).
Hint Constructors has_type : core.
Module Examples2.
Import Examples.
Do the following exercises after you have added product types to
the language. For each informal typing judgement, write it as a
formal statement in Coq and prove it.
Exercise: 1 star, standard, optional (typing_example_0)
(* empty ⊢ ((\z:A.z), (\z:B.z)) ∈ (A->A * B->B) *)
(* FILL IN HERE *)
☐
(* FILL IN HERE *)
☐
(* empty ⊢ (\x:(Top * B->B). x.snd) ((\z:A.z), (\z:B.z))
∈ B->B *)
(* FILL IN HERE *)
☐
∈ B->B *)
(* FILL IN HERE *)
☐
(* empty ⊢ (\z:(C->C)->(Top * B->B). (z (\x:C.x)).snd)
(\z:C->C. ((\z:A.z), (\z:B.z)))
∈ B->B *)
(* FILL IN HERE *)
☐
(\z:C->C. ((\z:A.z), (\z:B.z)))
∈ B->B *)
(* FILL IN HERE *)
☐
Properties
Inversion Lemmas for Subtyping
- Bool is the only subtype of Bool, and
- every subtype of an arrow type is itself an arrow type.
Exercise: 2 stars, standard, optional (sub_inversion_Bool)
Lemma sub_inversion_arrow : ∀ U V_{1} V_{2},
U <: <{V_{1}→V_{2}}> →
∃ U_{1} U_{2},
U = <{U_{1}→U_{2}}> ∧ V_{1} <: U_{1} ∧ U_{2} <: V_{2}.
U <: <{V_{1}→V_{2}}> →
∃ U_{1} U_{2},
U = <{U_{1}→U_{2}}> ∧ V_{1} <: U_{1} ∧ U_{2} <: V_{2}.
☐
Canonical Forms
Exercise: 3 stars, standard, optional (canonical_forms_of_arrow_types)
Lemma canonical_forms_of_arrow_types : ∀ Gamma s T_{1} T_{2},
Gamma ⊢ s \in (T_{1}→T_{2}) →
value s →
∃ x S_{1} s_{2},
s = <{\x:S_{1},s_{2}}>.
Gamma ⊢ s \in (T_{1}→T_{2}) →
value s →
∃ x S_{1} s_{2},
s = <{\x:S_{1},s_{2}}>.
Proof with eauto.
(* FILL IN HERE *) Admitted.
(* FILL IN HERE *) Admitted.
☐
Similarly, the canonical forms of type Bool are the constants
tm_true and tm_false.
Lemma canonical_forms_of_Bool : ∀ Gamma s,
Gamma ⊢ s \in Bool →
value s →
s = tm_true ∨ s = tm_false.
Gamma ⊢ s \in Bool →
value s →
s = tm_true ∨ s = tm_false.
Proof with eauto.
intros Gamma s Hty Hv.
remember <{Bool}> as T.
induction Hty; try solve_by_invert...
- (* T_Sub *)
subst. apply sub_inversion_Bool in H. subst...
Qed.
intros Gamma s Hty Hv.
remember <{Bool}> as T.
induction Hty; try solve_by_invert...
- (* T_Sub *)
subst. apply sub_inversion_Bool in H. subst...
Qed.
Progress
- If the last step in the typing derivation uses rule T_App,
then there are terms t_{1} t_{2} and types T_{1} and T_{2} such that
t = t_{1} t_{2}, T = T_{2}, empty ⊢ t_{1} \in T_{1} → T_{2}, and empty ⊢
t_{2} \in T_{1}. Moreover, by the induction hypothesis, either t_{1} is
a value or it steps, and either t_{2} is a value or it steps.
There are three possibilities to consider:
- Suppose t_{1} --> t_{1}' for some term t_{1}'. Then t_{1} t_{2} --> t_{1}' t_{2}
by ST_App1.
- Suppose t_{1} is a value and t_{2} --> t_{2}' for some term t_{2}'.
Then t_{1} t_{2} --> t_{1} t_{2}' by rule ST_App2 because t_{1} is a
value.
- Finally, suppose t_{1} and t_{2} are both values. By the
canonical forms lemma for arrow types, we know that t_{1} has the
form \x:S_{1}.s2 for some x, S_{1}, and s_{2}. But then
(\x:S_{1}.s2) t_{2} --> [x:=t_{2}]s_{2} by ST_AppAbs, since t_{2} is a
value.
- Suppose t_{1} --> t_{1}' for some term t_{1}'. Then t_{1} t_{2} --> t_{1}' t_{2}
by ST_App1.
- If the final step of the derivation uses rule T_Test, then there
are terms t_{1}, t_{2}, and t_{3} such that t = tm_if t_{1} then t_{2} else
t_{3}, with empty ⊢ t_{1} \in Bool and with empty ⊢ t_{2} \in T and
empty ⊢ t_{3} \in T. Moreover, by the induction hypothesis,
either t_{1} is a value or it steps.
- If t_{1} is a value, then by the canonical forms lemma for
booleans, either t_{1} = tm_true or t_{1} = tm_false. In either
case, t can step, using rule ST_TestTrue or ST_TestFalse.
- If t_{1} can step, then so can t, by rule ST_Test.
- If t_{1} is a value, then by the canonical forms lemma for
booleans, either t_{1} = tm_true or t_{1} = tm_false. In either
case, t can step, using rule ST_TestTrue or ST_TestFalse.
- If the final step of the derivation is by T_Sub, then there is a type T_{2} such that T_{1} <: T_{2} and empty ⊢ t_{1} \in T_{1}. The desired result is exactly the induction hypothesis for the typing subderivation.
Theorem progress : ∀ t T,
empty ⊢ t \in T →
value t ∨ ∃ t', t --> t'.
empty ⊢ t \in T →
value t ∨ ∃ t', t --> t'.
Proof with eauto.
intros t T Ht.
remember empty as Gamma.
induction Ht; subst Gamma; auto.
- (* T_Var *)
discriminate.
- (* T_App *)
right.
destruct IHHt1; subst...
+ (* t_{1} is a value *)
destruct IHHt2; subst...
× (* t_{2} is a value *)
eapply canonical_forms_of_arrow_types in Ht_{1}; [|assumption].
destruct Ht_{1} as [x [S_{1} [s_{2} H_{1}]]]. subst.
∃ (<{ [x:=t_{2}]s_{2} }>)...
× (* t_{2} steps *)
destruct H_{0} as [t_{2}' Hstp]. ∃ <{ t_{1} t_{2}' }>...
+ (* t_{1} steps *)
destruct H as [t_{1}' Hstp]. ∃ <{ t_{1}' t_{2} }>...
- (* T_Test *)
right.
destruct IHHt1.
+ (* t_{1} is a value *) eauto.
+ apply canonical_forms_of_Bool in Ht_{1}; [|assumption].
destruct Ht_{1}; subst...
+ destruct H. rename x into t_{1}'. eauto.
Qed.
intros t T Ht.
remember empty as Gamma.
induction Ht; subst Gamma; auto.
- (* T_Var *)
discriminate.
- (* T_App *)
right.
destruct IHHt1; subst...
+ (* t_{1} is a value *)
destruct IHHt2; subst...
× (* t_{2} is a value *)
eapply canonical_forms_of_arrow_types in Ht_{1}; [|assumption].
destruct Ht_{1} as [x [S_{1} [s_{2} H_{1}]]]. subst.
∃ (<{ [x:=t_{2}]s_{2} }>)...
× (* t_{2} steps *)
destruct H_{0} as [t_{2}' Hstp]. ∃ <{ t_{1} t_{2}' }>...
+ (* t_{1} steps *)
destruct H as [t_{1}' Hstp]. ∃ <{ t_{1}' t_{2} }>...
- (* T_Test *)
right.
destruct IHHt1.
+ (* t_{1} is a value *) eauto.
+ apply canonical_forms_of_Bool in Ht_{1}; [|assumption].
destruct Ht_{1}; subst...
+ destruct H. rename x into t_{1}'. eauto.
Qed.
Inversion Lemmas for Typing
- If the last step of the derivation is a use of T_Abs then there is a type T_{12} such that T = S_{1} → T_{12} and x:S_{1}; Gamma ⊢ t_{2} \in T_{12}. Picking T_{12} for S_{2} gives us what we need: S_{1} → T_{12} <: S_{1} → T_{12} follows from S_Refl.
- If the last step of the derivation is a use of T_Sub then there is a type S such that S <: T and Gamma ⊢ \x:S_{1}.t2 \in S. The IH for the typing subderivation tells us that there is some type S_{2} with S_{1} → S_{2} <: S and x:S_{1}; Gamma ⊢ t_{2} \in S_{2}. Picking type S_{2} gives us what we need, since S_{1} → S_{2} <: T then follows by S_Trans.
Lemma typing_inversion_abs : ∀ Gamma x S_{1} t_{2} T,
Gamma ⊢ \x:S_{1},t_{2} \in T →
∃ S_{2},
<{S_{1}→S_{2}}> <: T
∧ (x ⊢> S_{1} ; Gamma) ⊢ t_{2} \in S_{2}.
Gamma ⊢ \x:S_{1},t_{2} \in T →
∃ S_{2},
<{S_{1}→S_{2}}> <: T
∧ (x ⊢> S_{1} ; Gamma) ⊢ t_{2} \in S_{2}.
Lemma typing_inversion_var : ∀ Gamma (x:string) T,
Gamma ⊢ x \in T →
∃ S,
Gamma x = Some S ∧ S <: T.
Proof with eauto.
(* FILL IN HERE *) Admitted.
☐
Gamma ⊢ x \in T →
∃ S,
Gamma x = Some S ∧ S <: T.
Proof with eauto.
(* FILL IN HERE *) Admitted.
☐
Lemma typing_inversion_app : ∀ Gamma t_{1} t_{2} T_{2},
Gamma ⊢ t_{1} t_{2} \in T_{2} →
∃ T_{1},
Gamma ⊢ t_{1} \in (T_{1}→T_{2}) ∧
Gamma ⊢ t_{2} \in T_{1}.
Proof with eauto.
(* FILL IN HERE *) Admitted.
☐
Gamma ⊢ t_{1} t_{2} \in T_{2} →
∃ T_{1},
Gamma ⊢ t_{1} \in (T_{1}→T_{2}) ∧
Gamma ⊢ t_{2} \in T_{1}.
Proof with eauto.
(* FILL IN HERE *) Admitted.
☐
Lemma abs_arrow : ∀ x S_{1} s_{2} T_{1} T_{2},
empty ⊢ \x:S_{1},s_{2} \in (T_{1}→T_{2}) →
T_{1} <: S_{1}
∧ (x ⊢> S_{1} ; empty) ⊢ s_{2} \in T_{2}.
empty ⊢ \x:S_{1},s_{2} \in (T_{1}→T_{2}) →
T_{1} <: S_{1}
∧ (x ⊢> S_{1} ; empty) ⊢ s_{2} \in T_{2}.
Proof with eauto.
intros x S_{1} s_{2} T_{1} T_{2} Hty.
apply typing_inversion_abs in Hty.
destruct Hty as [S_{2} [Hsub Hty1]].
apply sub_inversion_arrow in Hsub.
destruct Hsub as [U_{1} [U_{2} [Heq [Hsub1 Hsub2]]]].
injection Heq as Heq; subst... Qed.
intros x S_{1} s_{2} T_{1} T_{2} Hty.
apply typing_inversion_abs in Hty.
destruct Hty as [S_{2} [Hsub Hty1]].
apply sub_inversion_arrow in Hsub.
destruct Hsub as [U_{1} [U_{2} [Heq [Hsub1 Hsub2]]]].
injection Heq as Heq; subst... Qed.
Lemma weakening : ∀ Gamma Gamma' t T,
inclusion Gamma Gamma' →
Gamma ⊢ t \in T →
Gamma' ⊢ t \in T.
Proof.
intros Gamma Gamma' t T H Ht.
generalize dependent Gamma'.
induction Ht; eauto using inclusion_update.
Qed.
Lemma weakening_empty : ∀ Gamma t T,
empty ⊢ t \in T →
Gamma ⊢ t \in T.
Proof.
intros Gamma t T.
eapply weakening.
discriminate.
Qed.
inclusion Gamma Gamma' →
Gamma ⊢ t \in T →
Gamma' ⊢ t \in T.
Proof.
intros Gamma Gamma' t T H Ht.
generalize dependent Gamma'.
induction Ht; eauto using inclusion_update.
Qed.
Lemma weakening_empty : ∀ Gamma t T,
empty ⊢ t \in T →
Gamma ⊢ t \in T.
Proof.
intros Gamma t T.
eapply weakening.
discriminate.
Qed.
Substitution
Lemma substitution_preserves_typing : ∀ Gamma x U t v T,
(x ⊢> U ; Gamma) ⊢ t \in T →
empty ⊢ v \in U →
Gamma ⊢ [x:=v]t \in T.
Proof.
(x ⊢> U ; Gamma) ⊢ t \in T →
empty ⊢ v \in U →
Gamma ⊢ [x:=v]t \in T.
Proof.
Proof.
intros Gamma x U t v T Ht Hv.
remember (x ⊢> U; Gamma) as Gamma'.
generalize dependent Gamma.
induction Ht; intros Gamma' G; simpl; eauto.
(* FILL IN HERE *) Admitted.
intros Gamma x U t v T Ht Hv.
remember (x ⊢> U; Gamma) as Gamma'.
generalize dependent Gamma.
induction Ht; intros Gamma' G; simpl; eauto.
(* FILL IN HERE *) Admitted.
Preservation
- If the final step of the derivation is by T_App, then there
are terms t_{1} and t_{2} and types T_{1} and T_{2} such that
t = t_{1} t_{2}, T = T_{2}, empty ⊢ t_{1} \in T_{1} → T_{2}, and
empty ⊢ t_{2} \in T_{1}.
- If the final step of the derivation uses rule T_Test, then
there are terms t_{1}, t_{2}, and t_{3} such that t = tm_if t_{1} then
t_{2} else t_{3}, with empty ⊢ t_{1} \in Bool and with empty ⊢ t_{2}
\in T and empty ⊢ t_{3} \in T. Moreover, by the induction
hypothesis, if t_{1} steps to t_{1}' then empty ⊢ t_{1}' : Bool.
There are three cases to consider, depending on which rule was
used to show t --> t'.
- If t --> t' by rule ST_Test, then t' = tm_if t_{1}' then t_{2}
else t_{3} with t_{1} --> t_{1}'. By the induction hypothesis,
empty ⊢ t_{1}' \in Bool, and so empty ⊢ t' \in T by
T_Test.
- If t --> t' by rule ST_TestTrue or ST_TestFalse, then
either t' = t_{2} or t' = t_{3}, and empty ⊢ t' \in T
follows by assumption.
- If t --> t' by rule ST_Test, then t' = tm_if t_{1}' then t_{2}
else t_{3} with t_{1} --> t_{1}'. By the induction hypothesis,
empty ⊢ t_{1}' \in Bool, and so empty ⊢ t' \in T by
T_Test.
- If the final step of the derivation uses rule T_Test, then
there are terms t_{1}, t_{2}, and t_{3} such that t = tm_if t_{1} then
t_{2} else t_{3}, with empty ⊢ t_{1} \in Bool and with empty ⊢ t_{2}
\in T and empty ⊢ t_{3} \in T. Moreover, by the induction
hypothesis, if t_{1} steps to t_{1}' then empty ⊢ t_{1}' : Bool.
There are three cases to consider, depending on which rule was
used to show t --> t'.
- If the final step of the derivation is by T_Sub, then there is a type S such that S <: T and empty ⊢ t \in S. The result is immediate by the induction hypothesis for the typing subderivation and an application of T_Sub. ☐
Theorem preservation : ∀ t t' T,
empty ⊢ t \in T →
t --> t' →
empty ⊢ t' \in T.
empty ⊢ t \in T →
t --> t' →
empty ⊢ t' \in T.
Proof with eauto.
intros t t' T HT. generalize dependent t'.
remember empty as Gamma.
induction HT;
intros t' HE; subst;
try solve [inversion HE; subst; eauto].
- (* T_App *)
inversion HE; subst...
(* Most of the cases are immediate by induction,
and eauto takes care of them *)
+ (* ST_AppAbs *)
destruct (abs_arrow _ _ _ _ _ HT_{1}) as [HA_{1} HA_{2}].
apply substitution_preserves_typing with T_{0}...
Qed.
intros t t' T HT. generalize dependent t'.
remember empty as Gamma.
induction HT;
intros t' HE; subst;
try solve [inversion HE; subst; eauto].
- (* T_App *)
inversion HE; subst...
(* Most of the cases are immediate by induction,
and eauto takes care of them *)
+ (* ST_AppAbs *)
destruct (abs_arrow _ _ _ _ _ HT_{1}) as [HA_{1} HA_{2}].
apply substitution_preserves_typing with T_{0}...
Qed.
Records, via Products and Top
{a:Nat, b:Nat} ----> {Nat,Nat} i.e., (Nat,(Nat,Top)) {c:Nat, a:Nat} ----> {Nat,Top,Nat} i.e., (Nat,(Top,(Nat,Top)))The encoding of record values doesn't change at all. It is easy (and instructive) to check that the subtyping rules above are validated by the encoding.
Exercises
Exercise: 2 stars, standard (variations)
Each part of this problem suggests a different way of changing the definition of the STLC with Unit and subtyping. (These changes are not cumulative: each part starts from the original language.) In each part, list which properties (Progress, Preservation, both, or neither) become false. If a property becomes false, give a counterexample.- Suppose we add the following typing rule:
Gamma ⊢ t ∈ S_{1}->S_{2} S_{1} <: T_{1} T_{1} <: S_{1} S_{2} <: T_{2} (T_Funny1) Gamma ⊢ t ∈ T_{1}->T_{2} - Suppose we add the following reduction rule:
(ST_Funny21) unit --> (\x:Top. x) - Suppose we add the following subtyping rule:
(S_Funny3) Unit <: Top->Top - Suppose we add the following subtyping rule:
(S_Funny4) Top->Top <: Unit - Suppose we add the following reduction rule:
(ST_Funny5) (unit t) --> (t unit) - Suppose we add the same reduction rule and a new typing rule:
(ST_Funny5) (unit t) --> (t unit) (T_Funny6) empty ⊢ unit ∈ Top->Top - Suppose we change the arrow subtyping rule to:
S_{1} <: T_{1} S_{2} <: T_{2} (S_Arrow') S_{1}->S_{2} <: T_{1}->T_{2}
(* Do not modify the following line: *)
Definition manual_grade_for_variations : option (nat×string) := None.
☐
Definition manual_grade_for_variations : option (nat×string) := None.
☐
Exercise: Adding Products
Exercise: 5 stars, standard (products)
Adding pairs, projections, and product types to the system we have defined is a relatively straightforward matter. Carry out this extension by modifying the definitions and proofs above:- Add constructors for pairs, first and second projections,
and product types to the definitions of ty and tm, and
extend the surrounding definitions accordingly
(refer to chapter MoreSTLC):
- value relation
- substitution
- operational semantics
- typing relation
- Extend the subtyping relation with this rule:
S_{1} <: T_{1} S_{2} <: T_{2} (S_Prod) S_{1} * S_{2} <: T_{1} * T_{2} - Extend the proofs of progress, preservation, and all their supporting lemmas to deal with the new constructs. (You'll also need to add a couple of completely new lemmas.)
(* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_products : option (nat×string) := None.
☐
(* Do not modify the following line: *)
Definition manual_grade_for_products : option (nat×string) := None.
☐
(* 2021-05-26 09:57 *)