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.

Concepts

We now turn to the study of subtyping, a key feature needed to support the object-oriented programming style.

A Motivating Example

Suppose we are writing a program involving two record types defined as follows:
      Person  = {name:String, age:Nat}
      Student = {name:String, age:Nat, gpa:Nat}
In the simply typed lamdba-calculus with records, the term
      (\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.
But this is silly: we're passing the function a better argument than it needs! The only thing the body of the function can possibly do with its record argument r is project the field age from it: nothing else is allowed by the type, and the presence or absence of an extra gpa field makes no difference at all. So, intuitively, it seems that this function should be applicable to any record value that has at least an age field.
More generally, a record with more fields is "at least as good in any context" as one with just a subset of these fields, in the sense that any value belonging to the longer record type can be used safely in any context expecting the shorter record type. If the context expects something with the shorter type but we actually give it something with the longer type, nothing bad will happen (formally, the program will not get stuck).
The principle at work here is called subtyping. We say that "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. The idea of subtyping applies not only to records, but to all of the type constructors in the language -- functions, pairs, etc.
Safe substitution principle:
  • 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

Subtyping plays a fundamental role in many programming languages -- in particular, it is closely related to the notion of subclassing in object-oriented languages.
An object in Java, C#, etc. can be thought of as a record, some of whose fields are functions ("methods") and some of whose fields are data values ("fields" or "instance variables"). Invoking a method m of an object o on some arguments a1..an roughly consists of projecting out the m field of o and applying it to a1..an.
The type of an object is called a class -- or, in some languages, an interface. It describes which methods and which data fields the object offers. Classes and interfaces are related by the subclass and subinterface relations. An object belonging to a subclass (or subinterface) is required to provide all the methods and fields of one belonging to a superclass (or superinterface), plus possibly some more.
The fact that an object from a subclass can be used in place of one from a superclass provides a degree of flexibility that is extremely handy for organizing complex libraries. For example, a GUI toolkit like Java's Swing framework might define an abstract interface Component that collects together the common fields and methods of all objects having a graphical representation that can be displayed on the screen and interact with the user, such as the buttons, checkboxes, and scrollbars of a typical GUI. A method that relies only on this common interface can now be applied to any of these objects.
Of course, real object-oriented languages include many other features besides these. For example, fields can be updated. Fields and methods can be declared "private". Classes can give initializers that are used when constructing objects. Code in subclasses can cooperate with code in superclasses via inheritance. Classes can have static methods and fields. Etc., etc.
To keep things simple here, we won't deal with any of these issues -- in fact, we won't even talk any more about objects or classes. (There is a lot of discussion in [Pierce 2002], if you are interested.) Instead, we'll study the core concepts behind the subclass / subinterface relation in the simplified setting of the STLC.

The Subsumption Rule

Our goal for this chapter is to add subtyping to the simply typed lambda-calculus (with some of the basic extensions from MoreStlc). This involves two steps:
  • Defining a binary subtype relation between types.
  • Enriching the typing relation to take subtyping into account.
The second step is actually very simple. We add just a single rule to the typing relation: the so-called rule of subsumption:
Gamma ⊢ t1 ∈ T1     T1 <: T2 (T_Sub)  

Gamma ⊢ t1 ∈ T2
This rule says, intuitively, that it is OK to "forget" some of what we know about a term.
For example, we may know that t1 is a record with two fields (e.g., T1 = {x:AA, y:BB}), but choose to forget about one of the fields (T2 = {y:BB}) so that we can pass t1 to a function that requires just a single-field record.

The Subtype Relation

The first step -- the definition of the relation S <: T -- is where all the action is. Let's look at each of the clauses of its definition.

Structural Rules

To start off, we impose two "structural rules" that are independent of any particular type constructor: a rule of transitivity, which says intuitively that, if S is better (richer, safer) than U and U is better than T, then S is better than T...
S <: U    U <: T (S_Trans)  

S <: T
... and a rule of reflexivity, since certainly any type T is as good as itself:
   (S_Refl)  

T <: T

Products

Now we consider the individual type constructors, one by one, beginning with product types. We consider one pair to be a subtype of another if each of its components is.
S1 <: T1    S2 <: T2 (S_Prod)  

S1 * S2 <: T1 * T2

Arrows

The subtyping rule for arrows is a little less intuitive. Suppose we have functions f and g with these types:
       f : CStudent
       g : (CPerson) → 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.
This example suggests that the subtyping rule for arrow types should say that two arrow types are in the subtype relation if their results are:
S2 <: T2 (S_Arrow_Co)  

S1 -> S2 <: S1 -> T2
We can generalize this to allow the arguments of the two arrow types to be in the subtype relation as well:
T1 <: S1    S2 <: T2 (S_Arrow)  

S1 -> S2 <: T1 -> T2
But notice that the argument types are subtypes "the other way round": in order to conclude that S1S2 to be a subtype of T1T2, it must be the case that T1 is a subtype of S1. The arrow constructor is said to be contravariant in its first argument and covariant in its second.
Here is an example that illustrates this:
       f : PersonC
       g : (StudentC) → 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.
The intuition is that, if we have a function f of type S1S2, then we know that f accepts elements of type S1; clearly, f will also accept elements of any subtype T1 of S1. The type of f also tells us that it returns elements of type S2; we can also view these results belonging to any supertype T2 of S2. That is, any function f of type S1S2 can also be viewed as having type T1T2.

Records

What about subtyping for record types?
The basic intuition is that it is always safe to use a "bigger" record in place of a "smaller" one. That is, given a record type, adding extra fields will always result in a subtype. If some code is expecting a record with fields x and y, it is perfectly safe for it to receive a record with fields x, y, and z; the z field will simply be ignored. For example,
    {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.
We can also create a subtype of a record type by replacing the type of one of its fields with a subtype. If some code is expecting a record with a field x of type T, it will be happy with a record having a field x of type S as long as S is a subtype of T. For example,
    {x:Student} <: {x:Person} This is known as "depth subtyping".
Finally, although the fields of a record type are written in a particular order, the order does not really matter. For example,
    {name:String,age:Nat} <: {age:Nat,name:String} This is known as "permutation subtyping".
We could formalize these requirements in a single subtyping rule for records as follows:
forall jk in j1..jn,
exists ip in i1..im, such that
jk=ip and Sp <: Tk (S_Rcd)  

{i1:S1...im:Sm} <: {j1:T1...jn:Tn}
That is, the record on the left should have all the field labels of the one on the right (and possibly more), while the types of the common fields should be in the subtype relation.
However, this rule is rather heavy and hard to read, so it is often decomposed into three simpler rules, which can be combined using S_Trans to achieve all the same effects.
First, adding fields to the end of a record type gives a subtype:
n > m (S_RcdWidth)  

{i1:T1...in:Tn} <: {i1:T1...im:Tm}
We can use S_RcdWidth to drop later fields of a multi-field record while keeping earlier fields, showing for example that {age:Nat,name:String} <: {age:Nat}.
Second, subtyping can be applied inside the components of a compound record type:
S1 <: T1 ... Sn <: Tn (S_RcdDepth)  

{i1:S1...in:Sn} <: {i1:T1...in:Tn}
For example, we can use S_RcdDepth and S_RcdWidth together to show that {y:Student, x:Nat} <: {y:Person}.
Third, subtyping can reorder fields. For example, we want {name:String, gpa:Nat, age:Nat} <: Person, but we haven't quite achieved this yet: using just S_RcdDepth and S_RcdWidth we can only drop fields from the end of a record type. So we add:
{i1:S1...in:Sn} is a permutation of {j1:T1...jn:Tn} (S_RcdPerm)  

{i1:S1...in:Sn} <: {j1:T1...jn:Tn}
It is worth noting that full-blown language designs may choose not to adopt all of these subtyping rules. For example, in Java:
  • 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:
S1 <: T1    S2 <: T2 (S_Arrow_wrong)  

S1 -> S2 <: T1 -> T2
Give a concrete example of functions f and g with the following types...
       f : StudentNat
       g : (PersonNat) → 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.

Top

Finally, it is convenient to give the subtype relation a maximum element -- a type that lies above every other type and is inhabited by all (well-typed) values. We do this by adding to the language one new type constant, called Top, together with a subtyping rule that places it above every other type in the subtype relation:
   (S_Top)  

S <: Top
The Top type is an analog of the Object type in Java and C#.

Summary

In summary, we form the STLC with subtyping by starting with the pure STLC (over some set of base types) and then...
  • adding a base type Top,
  • adding the rule of subsumption
    Gamma ⊢ t1 ∈ T1     T1 <: T2 (T_Sub)  

    Gamma ⊢ t1 ∈ T2
    to the typing relation, and
  • defining a subtype relation as follows:
    S <: U    U <: T (S_Trans)  

    S <: T
       (S_Refl)  

    T <: T
       (S_Top)  

    S <: Top
    S1 <: T1    S2 <: T2 (S_Prod)  

    S1 * S2 <: T1 * T2
    T1 <: S1    S2 <: T2 (S_Arrow)  

    S1 -> S2 <: T1 -> T2
    n > m (S_RcdWidth)  

    {i1:T1...in:Tn} <: {i1:T1...im:Tm}
    S1 <: T1 ... Sn <: Tn (S_RcdDepth)  

    {i1:S1...in:Sn} <: {i1:T1...in:Tn}
    {i1:S1...in:Sn} is a permutation of {j1:T1...jn:Tn} (S_RcdPerm)  

    {i1:S1...in:Sn} <: {j1:T1...jn:Tn}

Exercises

The following "thought exercises" are repeated later as formal 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.)
  • TS <: TS
  • TopU <: STop
  • (CC) (A×B) <: (CC) (Top×B)
  • TTU <: SSV
  • (TT)->U <: (SS)->V
  • ((TS)->T)->U <: ((ST)->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
Write these types in order from the most specific to the most general.
Where does the type TopTopStudent fit into this order? That is, state how Top (Top Student) compares with each of the five types above. It may be unrelated to some of them.
(* Do not modify the following line: *)
Definition manual_grade_for_subtype_order : option (nat×string) := None.

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
          SS <: TT

       S,
           S <: AA
            T,
              S = TTT <: A

       S T1 T2,
           (S <: T1T2) →
            S1 S2,
              S = S1S2T1 <: S1S2 <: T2

       S,
           S <: SS

       S,
           SS <: S

       S T1 T2,
           S <: T1×T2
            S1 S2,
              S = S1×S2S1 <: T1S2 <: T2
(* Do not modify the following line: *)
Definition manual_grade_for_subtype_instances_tf_2 : option (nat×string) := None.

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 S0, S1, 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 S0, S1, 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.

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 <: TST
(* Do not modify the following line: *)
Definition manual_grade_for_proper_subtypes : option (nat×string) := None.

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 AA
  • 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.

Exercise: 2 stars, standard (small_large_2)

  • What is the smallest type T that makes the following assertion true?
           empty ⊢ (\p:(AA × BB). 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.

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.

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.

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.

Exercise: 3 stars, standard, optional (count_supertypes)

How many supertypes does the record type {x:A, y:CC} have? That is, how many different types T are there such that {x:A, y:CC} <: 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 types
S1 <: T1    S2 <: T2 (S_Prod)  

S1*S2 <: T1*T2
intuitively corresponds to the "depth" subtyping rule for records. Extending the analogy, we might consider adding a "permutation" rule
    

T1*T2 <: T2*T1
for products. Is this a good idea? Briefly explain why or why not.
(* Do not modify the following line: *)
Definition manual_grade_for_pair_permutation : option (nat×string) := None.

Formal Definitions

Module STLCSub.
Most of the definitions needed to formalize what we've discussed above -- in particular, the syntax and operational semantics of the language -- are identical to what we saw in the last chapter. We just need to extend the typing relation with the subsumption rule and add a new Inductive definition for the subtyping relation. Let's first do the identical bits.
We include products in the syntax of types and terms, but not, for the moment, anywhere else; the products exercise below will ask you to extend the definitions of the value relation, operational semantics, subtyping relation, and typing relation and to extend the proofs of progress and preservation to fully support products.

Core Definitions

Syntax

In the rest of the chapter, we formalize just base types, booleans, arrow types, Unit, and Top, omitting record types and leaving product types as an exercise. For the sake of more interesting examples, we'll add an arbitrary set of base types like String, Float, etc. (Since they are just for examples, we won't bother adding any operations over these base types, but we could easily do so.)
Inductive ty : Type :=
  | Ty_Top : ty
  | Ty_Bool : ty
  | Ty_Base : string ty
  | Ty_Arrow : ty ty ty
  | Ty_Unit : ty
  | Ty_Prod : ty ty 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
  | tm_pair : tm tm tm
  | tm_fst : tm tm
  | tm_snd : tm 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).

Notation "X * Y" :=
  (Ty_Prod X Y) (in custom stlc at level 2, X custom stlc, Y custom stlc at level 0).
Notation "( x ',' y )" := (tm_pair x y) (in custom stlc at level 0,
                                                x custom stlc at level 99,
                                                y custom stlc at level 99).
Notation "t '.fst'" := (tm_fst t) (in custom stlc at level 0).
Notation "t '.snd'" := (tm_snd t) (in custom stlc at level 0).

Notation "{ x }" := x (in custom stlc at level 1, x constr).

Substitution

The definition of substitution remains exactly the same as for the pure STLC.
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 String.eqb x y then s else t
  | <{\y:T, t1}>
      if String.eqb x y then t else <{\y:T, [x:=s] t1}>
  | <{t1 t2}>
      <{([x:=s] t1) ([x:=s] t2)}>
  | <{true}>
      <{true}>
  | <{false}>
      <{false}>
  | <{if t1 then t2 else t3}>
      <{if ([x:=s] t1) then ([x:=s] t2) else ([x:=s] t3)}>
  | <{unit}>
      <{unit}>
  | <{ (t1, t2) }>
      <{( [x:=s] t1, [x:=s] t2 )}>
  | <{t0.fst}>
      <{ ([x:=s] t0).fst}>
  | <{t0.snd}>
      <{ ([x:=s] t0).snd}>
  end
where "'[' x ':=' s ']' t" := (subst x s t) (in custom stlc).

Reduction

Likewise the definitions of value and step.
Inductive value : tm Prop :=
  | v_abs : x T2 t1,
      value <{\x:T2, t1}>
  | 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 T2 t1 v2,
         value v2
         <{(\x:T2, t1) v2}> --> <{ [x:=v2]t1 }>
  | ST_App1 : t1 t1' t2,
         t1 --> t1'
         <{t1 t2}> --> <{t1' t2}>
  | ST_App2 : v1 t2 t2',
         value v1
         t2 --> t2'
         <{v1 t2}> --> <{v1 t2'}>
  | ST_IfTrue : t1 t2,
      <{if true then t1 else t2}> --> t1
  | ST_IfFalse : t1 t2,
      <{if false then t1 else t2}> --> t2
  | ST_If : t1 t1' t2 t3,
      t1 --> t1'
      <{if t1 then t2 else t3}> --> <{if t1' then t2 else t3}>
where "t '-->' t'" := (step t t').

Hint Constructors step : core.

Subtyping

Now we come to the interesting part. We begin by defining the subtyping relation and developing some of its important technical properties.
The definition of subtyping is just what we sketched in the motivating discussion.
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 : S1 S2 T1 T2,
      T1 <: S1
      S2 <: T2
      <{S1S2}> <: <{T1T2}>
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 :
  <{CBool}> <: <{CTop}>.
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).
Recall that, in chapter MoreStlc, the optional section "Encoding Records" describes how records can be encoded as pairs. Using this encoding, define pair types representing the following record types:
    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.
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.
The following facts are mostly easy to prove in Coq. To get full benefit from the exercises, make sure you also understand how to prove them on paper!

Exercise: 1 star, standard, optional (subtyping_example_1)

Example subtyping_example_1 :
  <{TopStudent}> <: <{(CC)Person}>.
Proof with eauto.
  (* FILL IN HERE *) Admitted.

Exercise: 1 star, standard, optional (subtyping_example_2)

Example subtyping_example_2 :
  <{TopPerson}> <: <{PersonTop}>.
Proof with eauto.
  (* FILL IN HERE *) Admitted.
End Examples.

Typing

The only change to the typing relation is the addition of the rule of subsumption, T_Sub.
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 T1,
      Gamma x = Some T1
      Gamma x \in T1
  | T_Abs : Gamma x T1 T2 t1,
      (x > T2 ; Gamma) t1 \in T1
      Gamma \x:T2, t1 \in (T2 T1)
  | T_App : T1 T2 Gamma t1 t2,
      Gamma t1 \in (T2 T1)
      Gamma t2 \in T2
      Gamma t1 t2 \in T1
  | T_True : Gamma,
       Gamma true \in Bool
  | T_False : Gamma,
       Gamma false \in Bool
  | T_If : t1 t2 t3 T1 Gamma,
       Gamma t1 \in Bool
       Gamma t2 \in T1
       Gamma t3 \in T1
       Gamma if t1 then t2 else t3 \in T1
  | T_Unit : Gamma,
      Gamma unit \in Unit
  (* New rule of subsumption: *)
  | T_Sub : Gamma t1 T1 T2,
      Gamma t1 \in T1
      T1 <: T2
      Gamma t1 \in T2

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 *)

Exercise: 2 stars, standard, optional (typing_example_1)

(* empty ⊢ (\x:(Top * B->B). x.snd) ((\z:A.z), (\z:B.z))
         ∈ B->B *)

(* FILL IN HERE *)

Exercise: 2 stars, standard, optional (typing_example_2)

(* 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 *)
End Examples2.

Properties

The fundamental properties of the system that we want to check are the same as always: progress and preservation. Unlike the extension of the STLC with references (chapter References), we don't need to change the statements of these properties to take subtyping into account. However, their proofs do become a little bit more involved.

Inversion Lemmas for Subtyping

Before we look at the properties of the typing relation, we need to establish a couple of critical structural properties of the subtype relation:
  • Bool is the only subtype of Bool, and
  • every subtype of an arrow type is itself an arrow type.
These are called inversion lemmas because they play a similar role in proofs as the built-in inversion tactic: given a hypothesis that there exists a derivation of some subtyping statement S <: T and some constraints on the shape of S and/or T, each inversion lemma reasons about what this derivation must look like to tell us something further about the shapes of S and T and the existence of subtype relations between their parts.

Exercise: 2 stars, standard, optional (sub_inversion_Bool)

Lemma sub_inversion_Bool : U,
     U <: <{Bool}>
     U = <{Bool}>.
Proof with auto.
  intros U Hs.
  remember <{Bool}> as V.
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars, standard (sub_inversion_arrow)

Lemma sub_inversion_arrow : U V1 V2,
     U <: <{V1V2}>
      U1 U2,
     U = <{U1U2}> V1 <: U1 U2 <: V2.
Proof with eauto.
  intros U V1 V2 Hs.
  remember <{V1V2}> as V.
  generalize dependent V2. generalize dependent V1.
  (* FILL IN HERE *) Admitted.
There are additional inversion lemmas for the other types:
  • Unit is the only subtype of Unit, and
  • Base n is the only subtype of Base n, and
  • Top is the only supertype of Top.

Exercise: 2 stars, standard, optional (sub_inversion_Unit)

Lemma sub_inversion_Unit : U,
     U <: <{Unit}>
     U = <{Unit}>.
Proof with auto.
  intros U Hs.
  remember <{Unit}> as V.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (sub_inversion_Base)

Lemma sub_inversion_Base : U s,
     U <: <{Base s}>
     U = <{Base s}>.
Proof with auto.
  intros U s Hs.
  remember <{Base s}> as V.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (sub_inversion_Top)

Lemma sub_inversion_Top : U,
     <{ Top }> <: U
     U = <{ Top }>.
Proof with auto.
  intros U Hs.
  remember <{Top}> as V.
  (* FILL IN HERE *) Admitted.

Canonical Forms

The proof of the progress theorem -- that a well-typed non-value can always take a step -- doesn't need to change too much: we just need one small refinement. When we're considering the case where the term in question is an application t1 t2 where both t1 and t2 are values, we need to know that t1 has the form of a lambda-abstraction, so that we can apply the ST_AppAbs reduction rule. In the ordinary STLC, this is obvious: we know that t1 has a function type T11T12, and there is only one rule that can be used to give a function type to a value -- rule T_Abs -- and the form of the conclusion of this rule forces t1 to be an abstraction.
In the STLC with subtyping, this reasoning doesn't quite work because there's another rule that can be used to show that a value has a function type: subsumption. Fortunately, this possibility doesn't change things much: if the last rule used to show Gamma t1 \in T11T12 is subsumption, then there is some sub-derivation whose subject is also t1, and we can reason by induction until we finally bottom out at a use of T_Abs.
This bit of reasoning is packaged up in the following lemma, which tells us the possible "canonical forms" (i.e., values) of function type.

Exercise: 3 stars, standard, optional (canonical_forms_of_arrow_types)

Lemma canonical_forms_of_arrow_types : Gamma s T1 T2,
  Gamma s \in (T1T2)
  value s
   x S1 s2,
     s = <{\x:S1,s2}>.
Proof with eauto.
  (* 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.
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.

Progress

The proof of progress now proceeds just like the one for the pure STLC, except that in several places we invoke canonical forms lemmas... Theorem (Progress): For any term t and type T, if empty t \in T then t is a value or t --> t' for some term t'.
Proof: Let t and T be given, with empty t \in T. Proceed by induction on the typing derivation.
The cases for T_Abs, T_Unit, T_True and T_False are immediate because abstractions, tm_unit, tm_true, and tm_false are already values. The T_Var case is vacuous because variables cannot be typed in the empty context. The remaining cases are more interesting:
  • If the last step in the typing derivation uses rule T_App, then there are terms t1 t2 and types T1 and T2 such that t = t1 t2, T = T2, empty t1 \in T1 T2, and empty t2 \in T1. Moreover, by the induction hypothesis, either t1 is a value or it steps, and either t2 is a value or it steps. There are three possibilities to consider:
    • Suppose t1 --> t1' for some term t1'. Then t1 t2 --> t1' t2 by ST_App1.
    • Suppose t1 is a value and t2 --> t2' for some term t2'. Then t1 t2 --> t1 t2' by rule ST_App2 because t1 is a value.
    • Finally, suppose t1 and t2 are both values. By the canonical forms lemma for arrow types, we know that t1 has the form \x:S1.s2 for some x, S1, and s2. But then (\x:S1.s2) t2 --> [x:=t2]s2 by ST_AppAbs, since t2 is a value.
  • If the final step of the derivation uses rule T_Test, then there are terms t1, t2, and t3 such that t = tm_if t1 then t2 else t3, with empty t1 \in Bool and with empty t2 \in T and empty t3 \in T. Moreover, by the induction hypothesis, either t1 is a value or it steps.
    • If t1 is a value, then by the canonical forms lemma for booleans, either t1 = tm_true or t1 = tm_false. In either case, t can step, using rule ST_TestTrue or ST_TestFalse.
    • If t1 can step, then so can t, by rule ST_Test.
  • If the final step of the derivation is by T_Sub, then there is a type T2 such that T1 <: T2 and empty t1 \in T1. The desired result is exactly the induction hypothesis for the typing subderivation.
Formally:
Theorem progress : 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...
    + (* t1 is a value *)
      destruct IHHt2; subst...
      × (* t2 is a value *)
        eapply canonical_forms_of_arrow_types in Ht1; [|assumption].
        destruct Ht1 as [x [S1 [s2 H1]]]. subst.
         (<{ [x:=t2]s2 }>)...
      × (* t2 steps *)
        destruct H0 as [t2' Hstp]. <{ t1 t2' }>...
    + (* t1 steps *)
      destruct H as [t1' Hstp]. <{ t1' t2 }>...
  - (* T_Test *)
    right.
    destruct IHHt1.
    + (* t1 is a value *) eauto.
    + apply canonical_forms_of_Bool in Ht1; [|assumption].
      destruct Ht1; subst...
    + destruct H. rename x into t1'. eauto.
Qed.

Inversion Lemmas for Typing

The proof of the preservation theorem also becomes a little more complex with the addition of subtyping. The reason is that, as with the "inversion lemmas for subtyping" above, there are a number of facts about the typing relation that are immediate from the definition in the pure STLC (formally: that can be obtained directly from the inversion tactic) but that require real proofs in the presence of subtyping because there are multiple ways to derive the same has_type statement.
The following inversion lemma tells us that, if we have a derivation of some typing statement Gamma \x:S1.t2 \in T whose subject is an abstraction, then there must be some subderivation giving a type to the body t2.
Lemma: If Gamma \x:S1.t2 \in T, then there is a type S2 such that x>S1; Gamma t2 \in S2 and S1 S2 <: T.
(Notice that the lemma does not say, "then T itself is an arrow type" -- this is tempting, but false!)
Proof: Let Gamma, x, S1, t2 and T be given as described. Proceed by induction on the derivation of Gamma \x:S1.t2 \in T. Cases T_Var, T_App, are vacuous as those rules cannot be used to give a type to a syntactic abstraction.
  • If the last step of the derivation is a use of T_Abs then there is a type T12 such that T = S1 T12 and x:S1; Gamma t2 \in T12. Picking T12 for S2 gives us what we need, since S1 T12 <: S1 T12 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:S1.t2 \in S. The IH for the typing subderivation tells us that there is some type S2 with S1 S2 <: S and x:S1; Gamma t2 \in S2. Picking type S2 gives us what we need, since S1 S2 <: T then follows by S_Trans.
Formally:
Lemma typing_inversion_abs : Gamma x S1 t2 T,
     Gamma \x:S1,t2 \in T
      S2,
       <{S1S2}> <: T
        (x > S1 ; Gamma) t2 \in S2.
Proof with eauto.
  intros Gamma x S1 t2 T H.
  remember <{\x:S1,t2}> as t.
  induction H;
    inversion Heqt; subst; intros; try solve_by_invert.
  - (* T_Abs *)
     T1...
  - (* T_Sub *)
    destruct IHhas_type as [S2 [Hsub Hty]]...
  Qed.

Exercise: 3 stars, standard, optional (typing_inversion_var)

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.

Exercise: 3 stars, standard, optional (typing_inversion_app)

Lemma typing_inversion_app : Gamma t1 t2 T2,
  Gamma t1 t2 \in T2
   T1,
    Gamma t1 \in (T1T2)
    Gamma t2 \in T1.
Proof with eauto.
  (* FILL IN HERE *) Admitted.

Lemma typing_inversion_unit : Gamma T,
  Gamma unit \in T
    <{Unit}> <: T.
Proof with eauto.
  intros Gamma T Htyp. remember <{ unit }> as tu.
  induction Htyp;
    inversion Heqtu; subst; intros...
Qed.

The inversion lemmas for typing and for subtyping between arrow types can be packaged up as a useful "combination lemma" telling us exactly what we'll actually require below.
Lemma abs_arrow : x S1 s2 T1 T2,
  empty \x:S1,s2 \in (T1T2)
     T1 <: S1
   (x > S1 ; empty) s2 \in T2.
Proof with eauto.
  intros x S1 s2 T1 T2 Hty.
  apply typing_inversion_abs in Hty.
  destruct Hty as [S2 [Hsub Hty1]].
  apply sub_inversion_arrow in Hsub.
  destruct Hsub as [U1 [U2 [Heq [Hsub1 Hsub2]]]].
  injection Heq as Heq; subst... Qed.

Weakening

The weakening lemma is proved as in pure STLC.
Lemma weakening : Gamma Gamma' t T,
     includedin 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 includedin_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

When subtyping is involved proofs are generally easier when done by induction on typing derivations, rather than on terms. The substitution lemma is proved as for pure STLC, but using induction on the typing derivation this time (see Exercise substitution_preserves_typing_from_typing_ind in StlcProp.v).
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.
  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

The proof of preservation now proceeds pretty much as in earlier chapters, using the substitution lemma at the appropriate point and the inversion lemma from above to extract structural information from typing assumptions.
Theorem (Preservation): If t, t' are terms and T is a type such that empty t \in T and t --> t', then empty t' \in T.
Proof: Let t and T be given such that empty t \in T. We proceed by induction on the structure of this typing derivation, leaving t' general. The cases T_Abs, T_Unit, T_True, and T_False cases are vacuous because abstractions and constants don't step. Case T_Var is vacuous as well, since the context is empty.
  • If the final step of the derivation is by T_App, then there are terms t1 and t2 and types T1 and T2 such that t = t1 t2, T = T2, empty t1 \in T1 T2, and empty t2 \in T1.
    By the definition of the step relation, there are three ways t1 t2 can step. Cases ST_App1 and ST_App2 follow immediately by the induction hypotheses for the typing subderivations and a use of T_App.
    Suppose instead t1 t2 steps by ST_AppAbs. Then t1 = \x:S.t12 for some type S and term t12, and t' = [x:=t2]t12.
    By lemma abs_arrow, we have T1 <: S and x:S1 s2 \in T2. It then follows by the substitution lemma (substitution_preserves_typing) that empty [x:=t2] t12 \in T2 as desired.
    • If the final step of the derivation uses rule T_Test, then there are terms t1, t2, and t3 such that t = tm_if t1 then t2 else t3, with empty t1 \in Bool and with empty t2 \in T and empty t3 \in T. Moreover, by the induction hypothesis, if t1 steps to t1' then empty t1' : 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 t1' then t2 else t3 with t1 --> t1'. By the induction hypothesis, empty t1' \in Bool, and so empty t' \in T by T_Test.
      • If t --> t' by rule ST_TestTrue or ST_TestFalse, then either t' = t2 or t' = t3, and empty t' \in T follows by assumption.
  • 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.
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 _ _ _ _ _ HT1) as [HA1 HA2].
      apply substitution_preserves_typing with T0...
Qed.

Records, via Products and Top

This formalization of the STLC with subtyping omits record types for brevity. If we want to deal with them more seriously, we have two choices.
First, we can treat them as part of the core language, writing down proper syntax, typing, and subtyping rules for them. Chapter RecordSub shows how this extension works.
On the other hand, if we are treating them as a derived form that is desugared in the parser, then we shouldn't need any new rules: we should just check that the existing rules for subtyping product and Unit types give rise to reasonable rules for record subtyping via this encoding. To do this, we just need to make one small change to the encoding described earlier: instead of using Unit as the base case in the encoding of tuples and the "don't care" placeholder in the encoding of records, we use Top. So:
    {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 ∈ S1->S2
    S1 <: T1     T1 <: S1      S2 <: T2 (T_Funny1)  

    Gamma ⊢ t ∈ T1->T2
  • 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:
    S1 <: T1 S2 <: T2 (S_Arrow')  

    S1->S2 <: T1->T2
(* Do not modify the following line: *)
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:
  • Constructors for pairs, first and second projections, and product types have already been added to the definitions of ty and tm. Also, the definition of substitution has been extended.
  • Extend the surrounding definitions accordingly (refer to chapter MoreSTLC):
    • value relation
    • operational semantics
    • typing relation
  • Extend the subtyping relation with this rule:
    S1 <: T1    S2 <: T2 (S_Prod)  

    S1 * S2 <: T1 * T2
  • 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_value_step : option (nat×string) := None.
(* Do not modify the following line: *)
Definition manual_grade_for_products_subtype_has_type : option (nat×string) := None.
(* Do not modify the following line: *)
Definition manual_grade_for_products_progress : option (nat×string) := None.
(* Do not modify the following line: *)
Definition manual_grade_for_products_preservation : option (nat×string) := None.

Formalized "Thought Exercises"

The following are formal exercises based on the previous "thought exercises."
Module FormalThoughtExercises.
Import Examples.
Notation p := "p".
Notation a := "a".

Definition TF P := P ¬P.

Exercise: 1 star, standard, optional (formal_subtype_instances_tf_1a)

Theorem formal_subtype_instances_tf_1a:
  TF ( S T U V, S <: T U <: V
         <{TS}> <: <{TS}>).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 1 star, standard, optional (formal_subtype_instances_tf_1b)

Theorem formal_subtype_instances_tf_1b:
  TF ( S T U V, S <: T U <: V
         <{TopU}> <: <{STop}>).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 1 star, standard, optional (formal_subtype_instances_tf_1c)

Theorem formal_subtype_instances_tf_1c:
  TF ( S T U V, S <: T U <: V
         <{(CC)(A×B)}> <: <{(CC)(Top×B)}>).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 1 star, standard, optional (formal_subtype_instances_tf_1d)

Theorem formal_subtype_instances_tf_1d:
  TF ( S T U V, S <: T U <: V
         <{T(TU)}> <: <{S(SV)}>).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 1 star, standard, optional (formal_subtype_instances_tf_1e)

Theorem formal_subtype_instances_tf_1e:
  TF ( S T U V, S <: T U <: V
         <{(TT)U}> <: <{(SS)V}>).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 1 star, standard, optional (formal_subtype_instances_tf_1f)

Theorem formal_subtype_instances_tf_1f:
  TF ( S T U V, S <: T U <: V
         <{((TS)T)U}> <: <{((ST)S)V}>).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 1 star, standard, optional (formal_subtype_instances_tf_1g)

Theorem formal_subtype_instances_tf_1g:
  TF ( S T U V, S <: T U <: V
         <{S×V}> <: <{T×U}>).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (formal_subtype_instances_tf_2a)

Theorem formal_subtype_instances_tf_2a:
  TF ( S T,
         S <: T
         <{SS}> <: <{TT}>).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (formal_subtype_instances_tf_2b)

Theorem formal_subtype_instances_tf_2b:
  TF ( S,
         S <: <{AA}>
          T,
           S = <{TT}> T <: A).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (formal_subtype_instances_tf_2d)

Hint: Assert a generalization of the statement to be proved and use induction on a type (rather than on a subtyping derviation).
Theorem formal_subtype_instances_tf_2d:
  TF ( S,
         S <: <{SS}>).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (formal_subtype_instances_tf_2e)

Theorem formal_subtype_instances_tf_2e:
  TF ( S,
         <{SS}> <: S).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (formal_subtype_concepts_tfa)

Theorem formal_subtype_concepts_tfa:
  TF ( T, S, S <: T).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (formal_subtype_concepts_tfb)

Theorem formal_subtype_concepts_tfb:
  TF ( T, S, T <: S).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (formal_subtype_concepts_tfc)

Theorem formal_subtype_concepts_tfc:
  TF ( T1 T2, S1 S2, <{S1×S2}> <: <{T1×T2}>).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (formal_subtype_concepts_tfd)

Theorem formal_subtype_concepts_tfd:
  TF ( T1 T2, S1 S2, <{T1×T2}> <: <{S1×S2}>).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (formal_subtype_concepts_tfe)

Theorem formal_subtype_concepts_tfe:
  TF ( T1 T2, S1 S2, <{S1S2}> <: <{T1T2}>).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (formal_subtype_concepts_tff)

Theorem formal_subtype_concepts_tff:
  TF ( T1 T2, S1 S2, <{T1T2}> <: <{S1S2}>).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (formal_subtype_concepts_tfg)

Theorem formal_subtype_concepts_tfg:
  TF ( f : nat ty,
         ( i j, i j f i f j)
         ( i, f (S i) <: f i)).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (formal_subtype_concepts_tfh)

Theorem formal_subtype_concepts_tfh:
  TF ( f : nat ty,
         ( i j, i j f i f j)
         ( i, f i <: f (S i))).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars, standard, optional (formal_proper_subtypes)

Theorem formal_proper_subtypes:
  TF ( T,
         ~(T = <{Bool}> ( n, T = <{Base n}>) T = <{Unit}>)
          S,
           S <: T S T).
Proof.
  (* FILL IN HERE *) Admitted.
Definition smallest_largest HT :=
  (* There exists a smallest and a largest. *)
  ( TS TL, T, TS <: T T <: TL HT T)
  
  (* There exists a smallest, but no largest. *)
  (( TS, T, TS <: T HT T)
   ~( TL, T, T <: TL HT T))
  
  (* There exists a largest, but not smallest. *)
  (~( TS, T, TS <: T HT T)
   ( TL, T, T <: TL HT T))
  
  (* There exists neither a smallest nor a largest. *)
  (~( TS, T, TS <: T HT T)
   ~( TL, T, T <: TL HT T)).

Exercise: 3 stars, advanced, optional (formal_small_large_1)

Theorem formal_small_large_1:
  smallest_largest
  (fun T
   empty <{(\p:T×Top, p.fst) ((\z:A, z), unit)}> \in <{AA}>).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars, advanced, optional (formal_small_large_2)

Theorem formal_small_large_2:
  smallest_largest
  (fun T
   empty <{(\p:(AA)×(BB), p) ((\z:A, z), (\z:B, z))}> \in T).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 4 stars, advanced, optional (formal_small_large_3)

Theorem formal_small_large_3:
  smallest_largest
  (fun T
   (a > A) <{(\p:A×T, (p.snd) (p.fst)) (a, \z:A, z)}> \in A).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 4 stars, advanced, optional (formal_small_large_4)

Theorem formal_small_large_4:
  smallest_largest
  (fun T
    S,
     empty <{\p:A×T, (p.snd) (p.fst)}> \in S).
Proof.
  (* FILL IN HERE *) Admitted.
Definition smallest P :=
  TF ( TS, T, TS <: T P T).

Exercise: 3 stars, standard, optional (formal_smallest_1)

Theorem formal_smallest_1:
  smallest
  (fun T
    S t,
     empty <{ (\x:T, x x) t }> \in S).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars, standard, optional (formal_smallest_2)

Theorem formal_smallest_2:
  smallest
  (fun T
   empty <{(\x:Top, x) ((\z:A, z), (\z:B, z))}> \in T).
Proof.
  (* FILL IN HERE *) Admitted.
End FormalThoughtExercises.

End STLCSub.

(* 2022-09-20 16:43 *)