SubSubtyping
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: it involves an application of a function that wants a one-field record to an argument that actually provides two fields, while the T_App rule demands that the domain type of the function being applied must match the type of the argument precisely.
Subtyping and Object-Oriented Languages
- mutable fields
- private and other visibility modifiers
- method inheritance
- static components
- etc., etc.
The Subsumption Rule
- Defining a binary subtype relation between types.
- Enriching the typing relation to take subtyping into account.
Γ ⊢ t : S S <: T | (T_Sub) |
Γ ⊢ t : T |
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 (function)
argument to yield a record of type Person. Also suppose, even
though we haven't yet discussed subtyping for records, 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.
g : (C→Person) → D
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.
g : (Student → C) → D
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.
{name:String, age:Nat} <: {name:String}
{name:String} <: {}
{x:Student} <: {x:Person}
This is known as "depth subtyping".
{name:String,age:Nat} <: {age:Nat,name:String}
This is known as "permutation subtyping".
for each jk in j_{1}..jn, | |
∃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 {i_{1}:T_{1}...in:Tn} | (S_RcdPerm) |
{i_{1}:S_{1}...in:Sn} <: {i_{1}:T_{1}...in:Tn} |
- A subclass may 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).
- Each class has just one superclass ("single inheritance" of
classes).
- 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).
Exercise: 2 stars, recommended (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.
g : (Person → Nat) → Nat
Top
(S_Top) | |
S <: Top |
Summary
- adding a base type Top,
- adding the rule of subsumption
to the typing relation, andΓ ⊢ t : S S <: T (T_Sub) Γ ⊢ t : T - 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 {i_{1}:T_{1}...in:Tn} (S_RcdPerm) {i_{1}:S_{1}...in:Sn} <: {i_{1}:T_{1}...in:Tn}
Exercises
Exercise: 1 star, 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.)- 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 (subtype_order)
The following types happen to form a linear order with respect to subtyping:- Top
- Top → Student
- Student → Person
- Student → Top
- Person → Student
Exercise: 1 star (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}
☐
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}
Exercise: 1 star (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.
Exercise: 2 stars (proper_subtypes)
Is the following statement true or false? Briefly explain your answer.
∀T,
~(∃n, T = TBase n) →
∃S,
S <: T ∧ S ≠ T
☐
~(∃n, T = TBase n) →
∃S,
S <: T ∧ S ≠ T
Exercise: 2 stars (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) : A→A
- What is the largest type T that makes the same assertion true?
Exercise: 2 stars (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)) : T
- What is the largest type T that makes the same assertion true?
Exercise: 2 stars, 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) : A
- What is the largest type T that makes the same assertion true?
Exercise: 2 stars (small_large_4)
- What is the smallest type T that makes the following
assertion true?
∃S,
empty ⊢ (\p:(A*T). (p.snd) (p.fst)) : S - What is the largest type T that makes the same assertion true?
Exercise: 2 stars (smallest_1)
What is the smallest type T that makes the following assertion true?
∃S, ∃t,
empty ⊢ (\x:T. x x) t : S
☐
empty ⊢ (\x:T. x x) t : S
Exercise: 2 stars (smallest_2)
What is the smallest type T that makes the following assertion true?
empty ⊢ (\x:Top. x) ((\z:A.z) , (\z:B.z)) : T
☐
Exercise: 3 stars, 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 (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} |
Formal Definitions
Syntax
Inductive ty : Type :=
| TTop : ty
| TBool : ty
| TBase : id → ty
| TArrow : ty → ty → ty
| TUnit : ty
.
Inductive tm : Type :=
| tvar : id → tm
| tapp : tm → tm → tm
| tabs : id → ty → tm → tm
| ttrue : tm
| tfalse : tm
| tif : tm → tm → tm → tm
| tunit : tm
.
Fixpoint subst (x:id) (s:tm) (t:tm) : tm :=
match t with
| tvar y ⇒
if beq_id x y then s else t
| tabs y T t_{1} ⇒
tabs y T (if beq_id x y then t_{1} else (subst x s t_{1}))
| tapp t_{1} t_{2} ⇒
tapp (subst x s t_{1}) (subst x s t_{2})
| ttrue ⇒
ttrue
| tfalse ⇒
tfalse
| tif t_{1} t_{2} t_{3} ⇒
tif (subst x s t_{1}) (subst x s t_{2}) (subst x s t_{3})
| tunit ⇒
tunit
end.
Notation "'[' x ':=' s ']' t" := (subst x s t) (at level 20).
Inductive value : tm → Prop :=
| v_abs : ∀x T t,
value (tabs x T t)
| v_true :
value ttrue
| v_false :
value tfalse
| v_unit :
value tunit
.
Hint Constructors value.
Reserved Notation "t_{1} '⇒' t_{2}" (at level 40).
Inductive step : tm → tm → Prop :=
| ST_AppAbs : ∀x T t_{12} v_{2},
value v_{2} →
(tapp (tabs x T t_{12}) v_{2}) ⇒ [x:=v_{2}]t_{12}
| ST_App1 : ∀t_{1} t_{1}' t_{2},
t_{1} ⇒ t_{1}' →
(tapp t_{1} t_{2}) ⇒ (tapp t_{1}' t_{2})
| ST_App2 : ∀v_{1} t_{2} t_{2}',
value v_{1} →
t_{2} ⇒ t_{2}' →
(tapp v_{1} t_{2}) ⇒ (tapp v_{1} t_{2}')
| ST_IfTrue : ∀t_{1} t_{2},
(tif ttrue t_{1} t_{2}) ⇒ t_{1}
| ST_IfFalse : ∀t_{1} t_{2},
(tif tfalse t_{1} t_{2}) ⇒ t_{2}
| ST_If : ∀t_{1} t_{1}' t_{2} t_{3},
t_{1} ⇒ t_{1}' →
(tif t_{1} t_{2} t_{3}) ⇒ (tif t_{1}' t_{2} t_{3})
where "t_{1} '⇒' t_{2}" := (step t_{1} t_{2}).
Hint Constructors step.
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 <: TTop
| S_Arrow : ∀S_{1} S_{2} T_{1} T_{2},
T_{1} <: S_{1} →
S_{2} <: T_{2} →
(TArrow S_{1} S_{2}) <: (TArrow T_{1} T_{2})
where "T '<:' U" := (subtype T U).
Note that we don't need any special rules for base types: they are
automatically subtypes of themselves (by S_Refl) and Top (by
S_Top), and that's all we want.
Hint Constructors subtype.
Module Examples.
Notation x := (Id 0).
Notation y := (Id 1).
Notation z := (Id 2).
Notation A := (TBase (Id 6)).
Notation B := (TBase (Id 7)).
Notation C := (TBase (Id 8)).
Notation String := (TBase (Id 9)).
Notation Float := (TBase (Id 10)).
Notation Integer := (TBase (Id 11)).
Exercise: 2 stars, optional (subtyping_judgements)
Person := { name : String }
Student := { name : String ;
gpa : Float }
Employee := { name : String ;
ssn : Integer }
Student := { name : String ;
gpa : Float }
Employee := { name : String ;
ssn : Integer }
Definition Person : ty :=
(* FILL IN HERE *) admit.
Definition Student : ty :=
(* FILL IN HERE *) admit.
Definition Employee : ty :=
(* FILL IN HERE *) admit.
Example sub_student_person :
Student <: Person.
Proof.
(* FILL IN HERE *) Admitted.
Example sub_employee_person :
Employee <: Person.
Proof.
(* FILL IN HERE *) Admitted.
☐
Example subtyping_example_0 :
(TArrow C Person) <: (TArrow C TTop).
(* C->Person <: C->Top *)
Proof.
apply S_Arrow.
apply S_Refl. auto.
Qed.
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, optional (subtyping_example_1)
Example subtyping_example_1 :
(TArrow TTop Student) <: (TArrow (TArrow C C) Person).
(* Top->Student <: (C->C)->Person *)
Proof with eauto.
(* FILL IN HERE *) Admitted.
(TArrow TTop Student) <: (TArrow (TArrow C C) Person).
(* Top->Student <: (C->C)->Person *)
Proof with eauto.
(* FILL IN HERE *) Admitted.
Example subtyping_example_2 :
(TArrow TTop Person) <: (TArrow Person TTop).
(* Top->Person <: Person->Top *)
Proof with eauto.
(* FILL IN HERE *) Admitted.
(TArrow TTop Person) <: (TArrow Person TTop).
(* Top->Person <: Person->Top *)
Proof with eauto.
(* FILL IN HERE *) Admitted.
☐
Definition context := partial_map ty.
Reserved Notation "Gamma '⊢' t '∈' T" (at level 40).
Inductive has_type : context → tm → ty → Prop :=
(* Same as before *)
| T_Var : ∀Γ x T,
Γ x = Some T →
Γ ⊢ (tvar x) ∈ T
| T_Abs : ∀Γ x T_{11} T_{12} t_{12},
(update Γ x T_{11}) ⊢ t_{12} ∈ T_{12} →
Γ ⊢ (tabs x T_{11} t_{12}) ∈ (TArrow T_{11} T_{12})
| T_App : ∀T_{1} T_{2} Γ t_{1} t_{2},
Γ ⊢ t_{1} ∈ (TArrow T_{1} T_{2}) →
Γ ⊢ t_{2} ∈ T_{1} →
Γ ⊢ (tapp t_{1} t_{2}) ∈ T_{2}
| T_True : ∀Γ,
Γ ⊢ ttrue ∈ TBool
| T_False : ∀Γ,
Γ ⊢ tfalse ∈ TBool
| T_If : ∀t_{1} t_{2} t_{3} T Γ,
Γ ⊢ t_{1} ∈ TBool →
Γ ⊢ t_{2} ∈ T →
Γ ⊢ t_{3} ∈ T →
Γ ⊢ (tif t_{1} t_{2} t_{3}) ∈ T
| T_Unit : ∀Γ,
Γ ⊢ tunit ∈ TUnit
(* New rule of subsumption *)
| T_Sub : ∀Γ t S T,
Γ ⊢ t ∈ S →
S <: T →
Γ ⊢ t ∈ T
where "Gamma '⊢' t '∈' T" := (has_type Γ t T).
Hint Constructors has_type.
(* To make your job simpler, the following hints help construct typing
derivations. *)
Hint Extern 2 (has_type _ (tapp _ _) _) ⇒
eapply T_App; auto.
Hint Extern 2 (_ = _) ⇒ compute; reflexivity.
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, optional (typing_example_0)
(* empty |- ((\z:A.z), (\z:B.z))
: (A->A * B->B) *)
(* FILL IN HERE *)
: (A->A * B->B) *)
(* 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
- every subtype of an arrow type is itself an arrow type.
Exercise: 2 stars, optional (sub_inversion_Bool)
Lemma sub_inversion_Bool : ∀U,
U <: TBool →
U = TBool.
Proof with auto.
intros U Hs.
remember TBool as V.
(* FILL IN HERE *) Admitted.
U <: TBool →
U = TBool.
Proof with auto.
intros U Hs.
remember TBool as V.
(* FILL IN HERE *) Admitted.
Lemma sub_inversion_arrow : ∀U V_{1} V_{2},
U <: (TArrow V_{1} V_{2}) →
∃U_{1}, ∃U_{2},
U = (TArrow U_{1} U_{2}) ∧ (V_{1} <: U_{1}) ∧ (U_{2} <: V_{2}).
Proof with eauto.
intros U V_{1} V_{2} Hs.
remember (TArrow V_{1} V_{2}) as V.
generalize dependent V_{2}. generalize dependent V_{1}.
(* FILL IN HERE *) Admitted.
U <: (TArrow V_{1} V_{2}) →
∃U_{1}, ∃U_{2},
U = (TArrow U_{1} U_{2}) ∧ (V_{1} <: U_{1}) ∧ (U_{2} <: V_{2}).
Proof with eauto.
intros U V_{1} V_{2} Hs.
remember (TArrow V_{1} V_{2}) as V.
generalize dependent V_{2}. generalize dependent V_{1}.
(* FILL IN HERE *) Admitted.
☐
Canonical Forms
Exercise: 3 stars, optional (canonical_forms_of_arrow_types)
Lemma canonical_forms_of_arrow_types : ∀Γ s T_{1} T_{2},
Γ ⊢ s ∈ (TArrow T_{1} T_{2}) →
value s →
∃x, ∃S_{1}, ∃s_{2},
s = tabs x S_{1} s_{2}.
Proof with eauto.
(* FILL IN HERE *) Admitted.
Γ ⊢ s ∈ (TArrow T_{1} T_{2}) →
value s →
∃x, ∃S_{1}, ∃s_{2},
s = tabs x S_{1} s_{2}.
Proof with eauto.
(* FILL IN HERE *) Admitted.
☐
Similarly, the canonical forms of type Bool are the constants
true and false.
Lemma canonical_forms_of_Bool : ∀Γ s,
Γ ⊢ s ∈ TBool →
value s →
(s = ttrue ∨ s = tfalse).
Proof with eauto.
intros Γ s Hty Hv.
remember TBool as T.
induction Hty; try solve by inversion...
- (* T_Sub *)
subst. apply sub_inversion_Bool in H. subst...
Qed.
intros Γ s Hty Hv.
remember TBool as T.
induction Hty; try solve by inversion...
- (* 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} : T_{1} → T_{2}, and empty ⊢
t_{2} : 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 Lemma
canonical_forms_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_If, then there
are terms t_{1}, t_{2}, and t_{3} such that t = if t_{1} then t_{2} else
t_{3}, with empty ⊢ t_{1} : Bool and with empty ⊢ t_{2} : T and
empty ⊢ t_{3} : 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} = true or t_{1} = false. In either
case, t can step, using rule ST_IfTrue or ST_IfFalse.
- If t_{1} can step, then so can t, by rule ST_If.
- If t_{1} is a value, then by the canonical forms lemma for
booleans, either t_{1} = true or t_{1} = false. In either
case, t can step, using rule ST_IfTrue or ST_IfFalse.
- If the final step of the derivation is by T_Sub, then there is a type S such that S <: T and empty ⊢ t : S. The desired result is exactly the induction hypothesis for the typing subderivation.
Theorem progress : ∀t T,
empty ⊢ t ∈ T →
value t ∨ ∃t', t ⇒ t'.
Proof with eauto.
intros t T Ht.
remember empty as Γ.
revert HeqGamma.
induction Ht;
intros HeqGamma; subst...
- (* T_Var *)
inversion H.
- (* T_App *)
right.
destruct IHHt1; subst...
+ (* t_{1} is a value *)
destruct IHHt2; subst...
* (* t_{2} is a value *)
destruct (canonical_forms_of_arrow_types empty t_{1} T_{1} T_{2})
as [x [S_{1} [t_{12} Heqt1]]]...
subst. ∃([x:=t_{2}]t_{12})...
* (* t_{2} steps *)
inversion H_{0} as [t_{2}' Hstp]. ∃(tapp t_{1} t_{2}')...
+ (* t_{1} steps *)
inversion H as [t_{1}' Hstp]. ∃(tapp t_{1}' t_{2})...
- (* T_If *)
right.
destruct IHHt1.
+ (* t_{1} is a value *) eauto.
+ assert (t_{1} = ttrue ∨ t_{1} = tfalse)
by (eapply canonical_forms_of_Bool; eauto).
inversion H_{0}; subst...
+ inversion H. rename x into t_{1}'. eauto.
Qed.
intros t T Ht.
remember empty as Γ.
revert HeqGamma.
induction Ht;
intros HeqGamma; subst...
- (* T_Var *)
inversion H.
- (* T_App *)
right.
destruct IHHt1; subst...
+ (* t_{1} is a value *)
destruct IHHt2; subst...
* (* t_{2} is a value *)
destruct (canonical_forms_of_arrow_types empty t_{1} T_{1} T_{2})
as [x [S_{1} [t_{12} Heqt1]]]...
subst. ∃([x:=t_{2}]t_{12})...
* (* t_{2} steps *)
inversion H_{0} as [t_{2}' Hstp]. ∃(tapp t_{1} t_{2}')...
+ (* t_{1} steps *)
inversion H as [t_{1}' Hstp]. ∃(tapp t_{1}' t_{2})...
- (* T_If *)
right.
destruct IHHt1.
+ (* t_{1} is a value *) eauto.
+ assert (t_{1} = ttrue ∨ t_{1} = tfalse)
by (eapply canonical_forms_of_Bool; eauto).
inversion H_{0}; subst...
+ inversion 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 Γ,
x:S_{1} ⊢ t_{2} : 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 Γ ⊢ \x:S_{1}.t2 : S. The IH for the typing subderivation tell us that there is some type S_{2} with S_{1} → S_{2} <: S and Γ, x:S_{1} ⊢ t_{2} : 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 : ∀Γ x S_{1} t_{2} T,
Γ ⊢ (tabs x S_{1} t_{2}) ∈ T →
(∃S_{2}, (TArrow S_{1} S_{2}) <: T
∧ (update Γ x S_{1}) ⊢ t_{2} ∈ S_{2}).
Proof with eauto.
intros Γ x S_{1} t_{2} T H.
remember (tabs x S_{1} t_{2}) as t.
induction H;
inversion Heqt; subst; intros; try solve by inversion.
- (* T_Abs *)
∃T_{12}...
- (* T_Sub *)
destruct IHhas_type as [S_{2} [Hsub Hty]]...
Qed.
intros Γ x S_{1} t_{2} T H.
remember (tabs x S_{1} t_{2}) as t.
induction H;
inversion Heqt; subst; intros; try solve by inversion.
- (* T_Abs *)
∃T_{12}...
- (* T_Sub *)
destruct IHhas_type as [S_{2} [Hsub Hty]]...
Qed.
Similarly...
Lemma typing_inversion_var : ∀Γ x T,
Γ ⊢ (tvar x) ∈ T →
∃S,
Γ x = Some S ∧ S <: T.
Proof with eauto.
intros Γ x T Hty.
remember (tvar x) as t.
induction Hty; intros;
inversion Heqt; subst; try solve by inversion.
- (* T_Var *)
∃T...
- (* T_Sub *)
destruct IHHty as [U [Hctx HsubU]]... Qed.
intros Γ x T Hty.
remember (tvar x) as t.
induction Hty; intros;
inversion Heqt; subst; try solve by inversion.
- (* T_Var *)
∃T...
- (* T_Sub *)
destruct IHHty as [U [Hctx HsubU]]... Qed.
Lemma typing_inversion_app : ∀Γ t_{1} t_{2} T_{2},
Γ ⊢ (tapp t_{1} t_{2}) ∈ T_{2} →
∃T_{1},
Γ ⊢ t_{1} ∈ (TArrow T_{1} T_{2}) ∧
Γ ⊢ t_{2} ∈ T_{1}.
Proof with eauto.
intros Γ t_{1} t_{2} T_{2} Hty.
remember (tapp t_{1} t_{2}) as t.
induction Hty; intros;
inversion Heqt; subst; try solve by inversion.
- (* T_App *)
∃T_{1}...
- (* T_Sub *)
destruct IHHty as [U_{1} [Hty1 Hty2]]...
Qed.
intros Γ t_{1} t_{2} T_{2} Hty.
remember (tapp t_{1} t_{2}) as t.
induction Hty; intros;
inversion Heqt; subst; try solve by inversion.
- (* T_App *)
∃T_{1}...
- (* T_Sub *)
destruct IHHty as [U_{1} [Hty1 Hty2]]...
Qed.
Lemma typing_inversion_true : ∀Γ T,
Γ ⊢ ttrue ∈ T →
TBool <: T.
Proof with eauto.
intros Γ T Htyp. remember ttrue as tu.
induction Htyp;
inversion Heqtu; subst; intros...
Qed.
intros Γ T Htyp. remember ttrue as tu.
induction Htyp;
inversion Heqtu; subst; intros...
Qed.
Lemma typing_inversion_false : ∀Γ T,
Γ ⊢ tfalse ∈ T →
TBool <: T.
Proof with eauto.
intros Γ T Htyp. remember tfalse as tu.
induction Htyp;
inversion Heqtu; subst; intros...
Qed.
intros Γ T Htyp. remember tfalse as tu.
induction Htyp;
inversion Heqtu; subst; intros...
Qed.
Lemma typing_inversion_if : ∀Γ t_{1} t_{2} t_{3} T,
Γ ⊢ (tif t_{1} t_{2} t_{3}) ∈ T →
Γ ⊢ t_{1} ∈ TBool
∧ Γ ⊢ t_{2} ∈ T
∧ Γ ⊢ t_{3} ∈ T.
Proof with eauto.
intros Γ t_{1} t_{2} t_{3} T Hty.
remember (tif t_{1} t_{2} t_{3}) as t.
induction Hty; intros;
inversion Heqt; subst; try solve by inversion.
- (* T_If *)
auto.
- (* T_Sub *)
destruct (IHHty H_{0}) as [H_{1} [H_{2} H_{3}]]...
Qed.
intros Γ t_{1} t_{2} t_{3} T Hty.
remember (tif t_{1} t_{2} t_{3}) as t.
induction Hty; intros;
inversion Heqt; subst; try solve by inversion.
- (* T_If *)
auto.
- (* T_Sub *)
destruct (IHHty H_{0}) as [H_{1} [H_{2} H_{3}]]...
Qed.
Lemma typing_inversion_unit : ∀Γ T,
Γ ⊢ tunit ∈ T →
TUnit <: T.
Proof with eauto.
intros Γ T Htyp. remember tunit as tu.
induction Htyp;
inversion Heqtu; subst; intros...
Qed.
intros Γ T Htyp. remember tunit 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 S_{1} s_{2} T_{1} T_{2},
empty ⊢ (tabs x S_{1} s_{2}) ∈ (TArrow T_{1} T_{2}) →
T_{1} <: S_{1}
∧ (update empty x S_{1}) ⊢ s_{2} ∈ T_{2}.
Proof with eauto.
intros x S_{1} s_{2} T_{1} T_{2} Hty.
apply typing_inversion_abs in Hty.
inversion Hty as [S_{2} [Hsub Hty1]].
apply sub_inversion_arrow in Hsub.
inversion Hsub as [U_{1} [U_{2} [Heq [Hsub1 Hsub2]]]].
inversion Heq; subst... Qed.
intros x S_{1} s_{2} T_{1} T_{2} Hty.
apply typing_inversion_abs in Hty.
inversion Hty as [S_{2} [Hsub Hty1]].
apply sub_inversion_arrow in Hsub.
inversion Hsub as [U_{1} [U_{2} [Heq [Hsub1 Hsub2]]]].
inversion Heq; subst... Qed.
Inductive appears_free_in : id → tm → Prop :=
| afi_var : ∀x,
appears_free_in x (tvar x)
| afi_app1 : ∀x t_{1} t_{2},
appears_free_in x t_{1} → appears_free_in x (tapp t_{1} t_{2})
| afi_app2 : ∀x t_{1} t_{2},
appears_free_in x t_{2} → appears_free_in x (tapp t_{1} t_{2})
| afi_abs : ∀x y T_{11} t_{12},
y ≠ x →
appears_free_in x t_{12} →
appears_free_in x (tabs y T_{11} t_{12})
| afi_if_{1} : ∀x t_{1} t_{2} t_{3},
appears_free_in x t_{1} →
appears_free_in x (tif t_{1} t_{2} t_{3})
| afi_if_{2} : ∀x t_{1} t_{2} t_{3},
appears_free_in x t_{2} →
appears_free_in x (tif t_{1} t_{2} t_{3})
| afi_if_{3} : ∀x t_{1} t_{2} t_{3},
appears_free_in x t_{3} →
appears_free_in x (tif t_{1} t_{2} t_{3})
.
Hint Constructors appears_free_in.
Lemma context_invariance : ∀Γ Γ' t S,
Γ ⊢ t ∈ S →
(∀x, appears_free_in x t → Γ x = Γ' x) →
Γ' ⊢ t ∈ S.
Proof with eauto.
intros. generalize dependent Γ'.
induction H;
intros Γ' Heqv...
- (* T_Var *)
apply T_Var... rewrite ← Heqv...
- (* T_Abs *)
apply T_Abs... apply IHhas_type. intros x_{0} Hafi.
unfold update, t_update. destruct (beq_idP x x_{0})...
- (* T_If *)
apply T_If...
Qed.
intros. generalize dependent Γ'.
induction H;
intros Γ' Heqv...
- (* T_Var *)
apply T_Var... rewrite ← Heqv...
- (* T_Abs *)
apply T_Abs... apply IHhas_type. intros x_{0} Hafi.
unfold update, t_update. destruct (beq_idP x x_{0})...
- (* T_If *)
apply T_If...
Qed.
Lemma free_in_context : ∀x t T Γ,
appears_free_in x t →
Γ ⊢ t ∈ T →
∃T', Γ x = Some T'.
Proof with eauto.
intros x t T Γ Hafi Htyp.
induction Htyp;
subst; inversion Hafi; subst...
- (* T_Abs *)
destruct (IHHtyp H_{4}) as [T Hctx]. ∃T.
unfold update, t_update in Hctx.
rewrite ← beq_id_false_iff in H_{2}.
rewrite H_{2} in Hctx... Qed.
intros x t T Γ Hafi Htyp.
induction Htyp;
subst; inversion Hafi; subst...
- (* T_Abs *)
destruct (IHHtyp H_{4}) as [T Hctx]. ∃T.
unfold update, t_update in Hctx.
rewrite ← beq_id_false_iff in H_{2}.
rewrite H_{2} in Hctx... Qed.
Substitution
Lemma substitution_preserves_typing : ∀Γ x U v t S,
(update Γ x U) ⊢ t ∈ S →
empty ⊢ v ∈ U →
Γ ⊢ ([x:=v]t) ∈ S.
Proof with eauto.
intros Γ x U v t S Htypt Htypv.
generalize dependent S. generalize dependent Γ.
induction t; intros; simpl.
- (* tvar *)
rename i into y.
destruct (typing_inversion_var _ _ _ Htypt)
as [T [Hctx Hsub]].
unfold update, t_update in Hctx.
destruct (beq_idP x y) as [Hxy|Hxy]; eauto;
subst.
inversion Hctx; subst. clear Hctx.
apply context_invariance with empty...
intros x Hcontra.
destruct (free_in_context _ _ S empty Hcontra)
as [T' HT']...
inversion HT'.
- (* tapp *)
destruct (typing_inversion_app _ _ _ _ Htypt)
as [T_{1} [Htypt1 Htypt2]].
eapply T_App...
- (* tabs *)
rename i into y. rename t into T_{1}.
destruct (typing_inversion_abs _ _ _ _ _ Htypt)
as [T_{2} [Hsub Htypt2]].
apply T_Sub with (TArrow T_{1} T_{2})... apply T_Abs...
destruct (beq_idP x y) as [Hxy|Hxy].
+ (* x=y *)
eapply context_invariance...
subst.
intros x Hafi. unfold update, t_update.
destruct (beq_id y x)...
+ (* x<>y *)
apply IHt. eapply context_invariance...
intros z Hafi. unfold update, t_update.
destruct (beq_idP y z)...
subst.
rewrite ← beq_id_false_iff in Hxy. rewrite Hxy...
- (* ttrue *)
assert (TBool <: S)
by apply (typing_inversion_true _ _ Htypt)...
- (* tfalse *)
assert (TBool <: S)
by apply (typing_inversion_false _ _ Htypt)...
- (* tif *)
assert ((update Γ x U) ⊢ t_{1} ∈ TBool
∧ (update Γ x U) ⊢ t_{2} ∈ S
∧ (update Γ x U) ⊢ t_{3} ∈ S)
by apply (typing_inversion_if _ _ _ _ _ Htypt).
inversion H as [H_{1} [H_{2} H_{3}]].
apply IHt1 in H_{1}. apply IHt2 in H_{2}. apply IHt3 in H_{3}.
auto.
- (* tunit *)
assert (TUnit <: S)
by apply (typing_inversion_unit _ _ Htypt)...
Qed.
intros Γ x U v t S Htypt Htypv.
generalize dependent S. generalize dependent Γ.
induction t; intros; simpl.
- (* tvar *)
rename i into y.
destruct (typing_inversion_var _ _ _ Htypt)
as [T [Hctx Hsub]].
unfold update, t_update in Hctx.
destruct (beq_idP x y) as [Hxy|Hxy]; eauto;
subst.
inversion Hctx; subst. clear Hctx.
apply context_invariance with empty...
intros x Hcontra.
destruct (free_in_context _ _ S empty Hcontra)
as [T' HT']...
inversion HT'.
- (* tapp *)
destruct (typing_inversion_app _ _ _ _ Htypt)
as [T_{1} [Htypt1 Htypt2]].
eapply T_App...
- (* tabs *)
rename i into y. rename t into T_{1}.
destruct (typing_inversion_abs _ _ _ _ _ Htypt)
as [T_{2} [Hsub Htypt2]].
apply T_Sub with (TArrow T_{1} T_{2})... apply T_Abs...
destruct (beq_idP x y) as [Hxy|Hxy].
+ (* x=y *)
eapply context_invariance...
subst.
intros x Hafi. unfold update, t_update.
destruct (beq_id y x)...
+ (* x<>y *)
apply IHt. eapply context_invariance...
intros z Hafi. unfold update, t_update.
destruct (beq_idP y z)...
subst.
rewrite ← beq_id_false_iff in Hxy. rewrite Hxy...
- (* ttrue *)
assert (TBool <: S)
by apply (typing_inversion_true _ _ Htypt)...
- (* tfalse *)
assert (TBool <: S)
by apply (typing_inversion_false _ _ Htypt)...
- (* tif *)
assert ((update Γ x U) ⊢ t_{1} ∈ TBool
∧ (update Γ x U) ⊢ t_{2} ∈ S
∧ (update Γ x U) ⊢ t_{3} ∈ S)
by apply (typing_inversion_if _ _ _ _ _ Htypt).
inversion H as [H_{1} [H_{2} H_{3}]].
apply IHt1 in H_{1}. apply IHt2 in H_{2}. apply IHt3 in H_{3}.
auto.
- (* tunit *)
assert (TUnit <: S)
by apply (typing_inversion_unit _ _ Htypt)...
Qed.
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} : T_{1} → T_{2}, and
empty ⊢ t_{2} : T_{1}.
- If the final step of the derivation uses rule T_If, then
there are terms t_{1}, t_{2}, and t_{3} such that t = if t_{1} then
t_{2} else t_{3}, with empty ⊢ t_{1} : Bool and with empty ⊢ t_{2} :
T and empty ⊢ t_{3} : 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_If, then t' = if t_{1}' then t_{2}
else t_{3} with t_{1} ⇒ t_{1}'. By the induction hypothesis,
empty ⊢ t_{1}' : Bool, and so empty ⊢ t' : T by T_If.
- If t ⇒ t' by rule ST_IfTrue or ST_IfFalse, then
either t' = t_{2} or t' = t_{3}, and empty ⊢ t' : T
follows by assumption.
- If t ⇒ t' by rule ST_If, then t' = if t_{1}' then t_{2}
else t_{3} with t_{1} ⇒ t_{1}'. By the induction hypothesis,
empty ⊢ t_{1}' : Bool, and so empty ⊢ t' : T by T_If.
- If the final step of the derivation uses rule T_If, then
there are terms t_{1}, t_{2}, and t_{3} such that t = if t_{1} then
t_{2} else t_{3}, with empty ⊢ t_{1} : Bool and with empty ⊢ t_{2} :
T and empty ⊢ t_{3} : 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 : 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 ∈ T →
t ⇒ t' →
empty ⊢ t' ∈ T.
Proof with eauto.
intros t t' T HT.
remember empty as Γ. generalize dependent HeqGamma.
generalize dependent t'.
induction HT;
intros t' HeqGamma HE; subst; inversion HE; subst...
- (* T_App *)
inversion HE; subst...
+ (* ST_AppAbs *)
destruct (abs_arrow _ _ _ _ _ HT_{1}) as [HA_{1} HA_{2}].
apply substitution_preserves_typing with T...
Qed.
intros t t' T HT.
remember empty as Γ. generalize dependent HeqGamma.
generalize dependent t'.
induction HT;
intros t' HeqGamma HE; subst; inversion HE; subst...
- (* T_App *)
inversion HE; subst...
+ (* ST_AppAbs *)
destruct (abs_arrow _ _ _ _ _ HT_{1}) as [HA_{1} HA_{2}].
apply substitution_preserves_typing with T...
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. For the rest of this chapter, we'll follow this encoding-based approach.
Exercises
Exercise: 2 stars (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:
Γ ⊢ t : S_{1}->S_{2} S_{1} <: T_{1} T_{1} <: S_{1} S_{2} <: T_{2} (T_Funny1) Γ ⊢ 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 evaluation rule:
(ST_Funny5) (unit t) ⇒ (t unit) - Suppose we add the same evaluation 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}
Exercise: Adding Products
Exercise: 4 stars (products)
Adding pairs, projections, and product types to the system we have defined is a relatively straightforward matter. Carry out this extension:- Add constructors for pairs, first and second projections, and
product types to the definitions of ty and tm. (Don't
forget to add corresponding cases to T_cases and t_cases.)
- Extend the substitution function and value relation as in
MoreSTLC.
- Extend the operational semantics with the same reduction rules
as in MoreSTLC.
- Extend the subtyping relation with this rule:
- —————————— (Sub_Prod)
S_{1} * S_{2} <: T_{1} * T_{2}
- —————————— (Sub_Prod)
S_{1} * S_{2} <: T_{1} * T_{2}
- Extend the typing relation with the same rules for pairs and
projections as in MoreSTLC.
- Extend the proofs of progress, preservation, and all their supporting lemmas to deal with the new constructs. (You'll also need to add some completely new lemmas.)