Library Sub
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
Of course, real OO languages have lots of other features...
- 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.
(T_Sub) Gamma |- t : T This rule says, intuitively, that it is OK to "forget" some of what we know about a term. For example, we may know that t is a record with two fields (e.g., S = {x:A→A, y:B→B}), but choose to forget about one of the fields (T = {y:B→B}) so that we can pass t to a function that requires just a single-field record.
The Subtype Relation
Structural Rules
(S_Trans) S <: T ... and a rule of reflexivity, since certainly any type T is as good as itself:
(S_Refl) T <: T
Products
(S_Prod) S1 * S2 <: T1 * T2
Arrows
(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 Notice that the argument types are subtypes "the other way round": in order to conclude that S1→S2 to be a subtype of T1→T2, 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.
Records
(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. If we like, we can decompose it into three simpler rules, which can be combined using S_Trans to achieve all the same effects.
(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} <: {name:String}.
(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}.
(S_RcdPerm) {i1:S1...in:Sn} <: {i1:T1...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 (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 : Student -> Nat g : (Person -> Nat) -> Nat ... such that the application g f will get stuck during execution.
Top
(S_Top) S <: Top The Top type is an analog of the Object type in Java and C#.
Summary
- adding a base type Top,
- adding the rule of subsumption
Gamma |- t : S S <: T
- ------------------------ (T_Sub) Gamma |- t : T
- defining a subtype relation as follows:
S <: U U <: T
- --------------- (S_Trans)
S <: T
- ----- (S_Refl)
- ------- (S_Top)
- ------------------- (S_Prod)
S1 * S2 <: T1 * T2
- ------------------- (S_Arrow)
- -------------------------------- (S_RcdWidth)
- --------------------------------- (S_RcdDepth)
- -------------------------------------------------- (S_RcdPerm) {i1:S1...in:Sn} <: {i1:T1...in:Tn}
- --------------- (S_Trans)
S <: T
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:Exercise: 1 star (subtype_instances_tf_2)
Which of the following statements are true? Write true or false after each one. forall S T, S <: T -> S->S <: T->TExercise: 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
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.
Exercise: 2 stars (proper_subtypes)
Is the following statement true or false? Briefly explain your answer. forall T, ~(exists n, T = TBase n) -> exists 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?
exists 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? exists S, exists t, 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 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.
Formal Definitions
Syntax
Inductive ty : Type :=
| TTop : ty
| TBool : ty
| TBase : id → ty
| TArrow : ty → ty → ty
| TUnit : ty
.
Tactic Notation "T_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "TTop" | Case_aux c "TBool"
| Case_aux c "TBase" | Case_aux c "TArrow"
| Case_aux c "TUnit" |
].
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
.
Tactic Notation "t_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "tvar" | Case_aux c "tapp"
| Case_aux c "tabs" | Case_aux c "ttrue"
| Case_aux c "tfalse" | Case_aux c "tif"
| Case_aux c "tunit"
].
Fixpoint subst (x:id) (s:tm) (t:tm) : tm :=
match t with
| tvar y ⇒
if eq_id_dec x y then s else t
| tabs y T t1 ⇒
tabs y T (if eq_id_dec x y then t1 else (subst x s t1))
| tapp t1 t2 ⇒
tapp (subst x s t1) (subst x s t2)
| ttrue ⇒
ttrue
| tfalse ⇒
tfalse
| tif t1 t2 t3 ⇒
tif (subst x s t1) (subst x s t2) (subst x s t3)
| 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 "t1 '==>' t2" (at level 40).
Inductive step : tm → tm → Prop :=
| ST_AppAbs : ∀ x T t12 v2,
value v2 →
(tapp (tabs x T t12) v2) ==> [x:=v2]t12
| ST_App1 : ∀ t1 t1' t2,
t1 ==> t1' →
(tapp t1 t2) ==> (tapp t1' t2)
| ST_App2 : ∀ v1 t2 t2',
value v1 →
t2 ==> t2' →
(tapp v1 t2) ==> (tapp v1 t2')
| ST_IfTrue : ∀ t1 t2,
(tif ttrue t1 t2) ==> t1
| ST_IfFalse : ∀ t1 t2,
(tif tfalse t1 t2) ==> t2
| ST_If : ∀ t1 t1' t2 t3,
t1 ==> t1' →
(tif t1 t2 t3) ==> (tif t1' t2 t3)
where "t1 '==>' t2" := (step t1 t2).
Tactic Notation "step_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "ST_AppAbs" | Case_aux c "ST_App1"
| Case_aux c "ST_App2" | Case_aux c "ST_IfTrue"
| Case_aux c "ST_IfFalse" | Case_aux c "ST_If"
].
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 : ∀ S1 S2 T1 T2,
T1 <: S1 →
S2 <: T2 →
(TArrow S1 S2) <: (TArrow T1 T2)
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.
Tactic Notation "subtype_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "S_Refl" | Case_aux c "S_Trans"
| Case_aux c "S_Top" | Case_aux c "S_Arrow"
].
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)
Definition Person : ty :=
admit.
Definition Student : ty :=
admit.
Definition Employee : ty :=
admit.
Example sub_student_person :
Student <: Person.
Proof.
Admitted.
Example sub_employee_person :
Employee <: Person.
Proof.
Admitted.
☐
Example subtyping_example_0 :
(TArrow C Person) <: (TArrow C TTop).
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).
Proof with eauto.
Admitted.
(TArrow TTop Student) <: (TArrow (TArrow C C) Person).
Proof with eauto.
Admitted.
Example subtyping_example_2 :
(TArrow TTop Person) <: (TArrow Person TTop).
Proof with eauto.
Admitted.
(TArrow TTop Person) <: (TArrow Person TTop).
Proof with eauto.
Admitted.
☐
Definition context := id → (option ty).
Definition empty : context := (fun _ ⇒ None).
Definition extend (Gamma : context) (x:id) (T : ty) :=
fun x' ⇒ if eq_id_dec x x' then Some T else Gamma x'.
Reserved Notation "Gamma '|-' t '\in' T" (at level 40).
Inductive has_type : context → tm → ty → Prop :=
| T_Var : ∀ Gamma x T,
Gamma x = Some T →
Gamma |- (tvar x) \in T
| T_Abs : ∀ Gamma x T11 T12 t12,
(extend Gamma x T11) |- t12 \in T12 →
Gamma |- (tabs x T11 t12) \in (TArrow T11 T12)
| T_App : ∀ T1 T2 Gamma t1 t2,
Gamma |- t1 \in (TArrow T1 T2) →
Gamma |- t2 \in T1 →
Gamma |- (tapp t1 t2) \in T2
| T_True : ∀ Gamma,
Gamma |- ttrue \in TBool
| T_False : ∀ Gamma,
Gamma |- tfalse \in TBool
| T_If : ∀ t1 t2 t3 T Gamma,
Gamma |- t1 \in TBool →
Gamma |- t2 \in T →
Gamma |- t3 \in T →
Gamma |- (tif t1 t2 t3) \in T
| T_Unit : ∀ Gamma,
Gamma |- tunit \in TUnit
| T_Sub : ∀ Gamma t S T,
Gamma |- t \in S →
S <: T →
Gamma |- t \in T
where "Gamma '|-' t '\in' T" := (has_type Gamma t T).
Hint Constructors has_type.
Tactic Notation "has_type_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "T_Var" | Case_aux c "T_Abs"
| Case_aux c "T_App" | Case_aux c "T_True"
| Case_aux c "T_False" | Case_aux c "T_If"
| Case_aux c "T_Unit"
| Case_aux c "T_Sub" ].
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)
☐
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.
Admitted.
U <: TBool →
U = TBool.
Proof with auto.
intros U Hs.
remember TBool as V.
Admitted.
Lemma sub_inversion_arrow : ∀ U V1 V2,
U <: (TArrow V1 V2) →
∃ U1, ∃ U2,
U = (TArrow U1 U2) ∧ (V1 <: U1) ∧ (U2 <: V2).
Proof with eauto.
intros U V1 V2 Hs.
remember (TArrow V1 V2) as V.
generalize dependent V2. generalize dependent V1.
Admitted.
U <: (TArrow V1 V2) →
∃ U1, ∃ U2,
U = (TArrow U1 U2) ∧ (V1 <: U1) ∧ (U2 <: V2).
Proof with eauto.
intros U V1 V2 Hs.
remember (TArrow V1 V2) as V.
generalize dependent V2. generalize dependent V1.
Admitted.
☐
Canonical Forms
Exercise: 3 stars, optional (canonical_forms_of_arrow_types)
Lemma canonical_forms_of_arrow_types : ∀ Gamma s T1 T2,
Gamma |- s \in (TArrow T1 T2) →
value s →
∃ x, ∃ S1, ∃ s2,
s = tabs x S1 s2.
Proof with eauto.
Admitted.
Gamma |- s \in (TArrow T1 T2) →
value s →
∃ x, ∃ S1, ∃ s2,
s = tabs x S1 s2.
Proof with eauto.
Admitted.
Lemma canonical_forms_of_Bool : ∀ Gamma s,
Gamma |- s \in TBool →
value s →
(s = ttrue ∨ s = tfalse).
Proof with eauto.
intros Gamma s Hty Hv.
remember TBool as T.
has_type_cases (induction Hty) Case; try solve by inversion...
Case "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 t1 t2 and types T1 and T2 such that
t = t1 t2, T = T2, empty |- t1 : T1 → T2, and empty |-
t2 : 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 Lemma
canonical_forms_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.
- Suppose t1 ==> t1' for some term t1'. Then t1 t2 ==> t1' t2
by ST_App1.
- If the final step of the derivation uses rule T_If, then there
are terms t1, t2, and t3 such that t = if t1 then t2 else
t3, with empty |- t1 : Bool and with empty |- t2 : T and
empty |- t3 : T. Moreover, by the induction hypothesis,
either t1 is a value or it steps.
- 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 \in T →
value t ∨ ∃ t', t ==> t'.
Proof with eauto.
intros t T Ht.
remember empty as Gamma.
revert HeqGamma.
has_type_cases (induction Ht) Case;
intros HeqGamma; subst...
Case "T_Var".
inversion H.
Case "T_App".
right.
destruct IHHt1; subst...
SCase "t1 is a value".
destruct IHHt2; subst...
SSCase "t2 is a value".
destruct (canonical_forms_of_arrow_types empty t1 T1 T2)
as [x [S1 [t12 Heqt1]]]...
subst. ∃ ([x:=t2]t12)...
SSCase "t2 steps".
inversion H0 as [t2' Hstp]. ∃ (tapp t1 t2')...
SCase "t1 steps".
inversion H as [t1' Hstp]. ∃ (tapp t1' t2)...
Case "T_If".
right.
destruct IHHt1.
SCase "t1 is a value"...
assert (t1 = ttrue ∨ t1 = tfalse)
by (eapply canonical_forms_of_Bool; eauto).
inversion H0; subst...
inversion H. rename x into t1'. eauto.
Qed.
Inversion Lemmas for Typing
- 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 Gamma,
x:S1 |- t2 : T12. Picking T12 for S2 gives us what we
need: 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 : S. The IH for the typing subderivation tell us that there is some type S2 with S1 → S2 <: S and Gamma, x:S1 |- t2 : S2. Picking type S2 gives us what we need, since S1 → S2 <: T then follows by S_Trans.
Lemma typing_inversion_abs : ∀ Gamma x S1 t2 T,
Gamma |- (tabs x S1 t2) \in T →
(∃ S2, (TArrow S1 S2) <: T
∧ (extend Gamma x S1) |- t2 \in S2).
Proof with eauto.
intros Gamma x S1 t2 T H.
remember (tabs x S1 t2) as t.
has_type_cases (induction H) Case;
inversion Heqt; subst; intros; try solve by inversion.
Case "T_Abs".
∃ T12...
Case "T_Sub".
destruct IHhas_type as [S2 [Hsub Hty]]...
Qed.
Similarly...
Lemma typing_inversion_var : ∀ Gamma x T,
Gamma |- (tvar x) \in T →
∃ S,
Gamma x = Some S ∧ S <: T.
Proof with eauto.
intros Gamma x T Hty.
remember (tvar x) as t.
has_type_cases (induction Hty) Case; intros;
inversion Heqt; subst; try solve by inversion.
Case "T_Var".
∃ T...
Case "T_Sub".
destruct IHHty as [U [Hctx HsubU]]... Qed.
Lemma typing_inversion_app : ∀ Gamma t1 t2 T2,
Gamma |- (tapp t1 t2) \in T2 →
∃ T1,
Gamma |- t1 \in (TArrow T1 T2) ∧
Gamma |- t2 \in T1.
Proof with eauto.
intros Gamma t1 t2 T2 Hty.
remember (tapp t1 t2) as t.
has_type_cases (induction Hty) Case; intros;
inversion Heqt; subst; try solve by inversion.
Case "T_App".
∃ T1...
Case "T_Sub".
destruct IHHty as [U1 [Hty1 Hty2]]...
Qed.
Lemma typing_inversion_true : ∀ Gamma T,
Gamma |- ttrue \in T →
TBool <: T.
Proof with eauto.
intros Gamma T Htyp. remember ttrue as tu.
has_type_cases (induction Htyp) Case;
inversion Heqtu; subst; intros...
Qed.
Lemma typing_inversion_false : ∀ Gamma T,
Gamma |- tfalse \in T →
TBool <: T.
Proof with eauto.
intros Gamma T Htyp. remember tfalse as tu.
has_type_cases (induction Htyp) Case;
inversion Heqtu; subst; intros...
Qed.
Lemma typing_inversion_if : ∀ Gamma t1 t2 t3 T,
Gamma |- (tif t1 t2 t3) \in T →
Gamma |- t1 \in TBool
∧ Gamma |- t2 \in T
∧ Gamma |- t3 \in T.
Proof with eauto.
intros Gamma t1 t2 t3 T Hty.
remember (tif t1 t2 t3) as t.
has_type_cases (induction Hty) Case; intros;
inversion Heqt; subst; try solve by inversion.
Case "T_If".
auto.
Case "T_Sub".
destruct (IHHty H0) as [H1 [H2 H3]]...
Qed.
Lemma typing_inversion_unit : ∀ Gamma T,
Gamma |- tunit \in T →
TUnit <: T.
Proof with eauto.
intros Gamma T Htyp. remember tunit as tu.
has_type_cases (induction Htyp) Case;
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 |- (tabs x S1 s2) \in (TArrow T1 T2) →
T1 <: S1
∧ (extend empty x S1) |- s2 \in T2.
Proof with eauto.
intros x S1 s2 T1 T2 Hty.
apply typing_inversion_abs in Hty.
inversion Hty as [S2 [Hsub Hty1]].
apply sub_inversion_arrow in Hsub.
inversion Hsub as [U1 [U2 [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 t1 t2,
appears_free_in x t1 → appears_free_in x (tapp t1 t2)
| afi_app2 : ∀ x t1 t2,
appears_free_in x t2 → appears_free_in x (tapp t1 t2)
| afi_abs : ∀ x y T11 t12,
y ≠ x →
appears_free_in x t12 →
appears_free_in x (tabs y T11 t12)
| afi_if1 : ∀ x t1 t2 t3,
appears_free_in x t1 →
appears_free_in x (tif t1 t2 t3)
| afi_if2 : ∀ x t1 t2 t3,
appears_free_in x t2 →
appears_free_in x (tif t1 t2 t3)
| afi_if3 : ∀ x t1 t2 t3,
appears_free_in x t3 →
appears_free_in x (tif t1 t2 t3)
.
Hint Constructors appears_free_in.
Lemma context_invariance : ∀ Gamma Gamma' t S,
Gamma |- t \in S →
(∀ x, appears_free_in x t → Gamma x = Gamma' x) →
Gamma' |- t \in S.
Proof with eauto.
intros. generalize dependent Gamma'.
has_type_cases (induction H) Case;
intros Gamma' Heqv...
Case "T_Var".
apply T_Var... rewrite <- Heqv...
Case "T_Abs".
apply T_Abs... apply IHhas_type. intros x0 Hafi.
unfold extend. destruct (eq_id_dec x x0)...
Case "T_App".
apply T_App with T1...
Case "T_If".
apply T_If...
Qed.
Lemma free_in_context : ∀ x t T Gamma,
appears_free_in x t →
Gamma |- t \in T →
∃ T', Gamma x = Some T'.
Proof with eauto.
intros x t T Gamma Hafi Htyp.
has_type_cases (induction Htyp) Case;
subst; inversion Hafi; subst...
Case "T_Abs".
destruct (IHHtyp H4) as [T Hctx]. ∃ T.
unfold extend in Hctx. rewrite neq_id in Hctx... Qed.
Substitution
Lemma substitution_preserves_typing : ∀ Gamma x U v t S,
(extend Gamma x U) |- t \in S →
empty |- v \in U →
Gamma |- ([x:=v]t) \in S.
Proof with eauto.
intros Gamma x U v t S Htypt Htypv.
generalize dependent S. generalize dependent Gamma.
t_cases (induction t) Case; intros; simpl.
Case "tvar".
rename i into y.
destruct (typing_inversion_var _ _ _ Htypt)
as [T [Hctx Hsub]].
unfold extend in Hctx.
destruct (eq_id_dec x y)...
SCase "x=y".
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'.
Case "tapp".
destruct (typing_inversion_app _ _ _ _ Htypt)
as [T1 [Htypt1 Htypt2]].
eapply T_App...
Case "tabs".
rename i into y. rename t into T1.
destruct (typing_inversion_abs _ _ _ _ _ Htypt)
as [T2 [Hsub Htypt2]].
apply T_Sub with (TArrow T1 T2)... apply T_Abs...
destruct (eq_id_dec x y).
SCase "x=y".
eapply context_invariance...
subst.
intros x Hafi. unfold extend.
destruct (eq_id_dec y x)...
SCase "x<>y".
apply IHt. eapply context_invariance...
intros z Hafi. unfold extend.
destruct (eq_id_dec y z)...
subst. rewrite neq_id...
Case "ttrue".
assert (TBool <: S)
by apply (typing_inversion_true _ _ Htypt)...
Case "tfalse".
assert (TBool <: S)
by apply (typing_inversion_false _ _ Htypt)...
Case "tif".
assert ((extend Gamma x U) |- t1 \in TBool
∧ (extend Gamma x U) |- t2 \in S
∧ (extend Gamma x U) |- t3 \in S)
by apply (typing_inversion_if _ _ _ _ _ Htypt).
inversion H as [H1 [H2 H3]].
apply IHt1 in H1. apply IHt2 in H2. apply IHt3 in H3.
auto.
Case "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 t1 and t2 and types T1 and T2 such that
t = t1 t2, T = T2, empty |- t1 : T1 → T2, and
empty |- t2 : T1.
- If the final step of the derivation uses rule T_If, then
there are terms t1, t2, and t3 such that t = if t1 then
t2 else t3, with empty |- t1 : Bool and with empty |- t2 :
T and empty |- t3 : 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 the final step of the derivation uses rule T_If, then
there are terms t1, t2, and t3 such that t = if t1 then
t2 else t3, with empty |- t1 : Bool and with empty |- t2 :
T and empty |- t3 : 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 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 \in T →
t ==> t' →
empty |- t' \in T.
Proof with eauto.
intros t t' T HT.
remember empty as Gamma. generalize dependent HeqGamma.
generalize dependent t'.
has_type_cases (induction HT) Case;
intros t' HeqGamma HE; subst; inversion HE; subst...
Case "T_App".
inversion HE; subst...
SCase "ST_AppAbs".
destruct (abs_arrow _ _ _ _ _ HT1) as [HA1 HA2].
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:
Gamma |- t : S1->S2
S1 <: T1 T1 <: S1 S2 <: T2
- ---------------------------------- (T_Funny1)
Gamma |- t : T1->T2
- ---------------------------------- (T_Funny1)
Gamma |- t : T1->T2
- Suppose we add the following reduction rule:
- ----------------- (ST_Funny21)
- Suppose we add the following subtyping rule:
- ------------- (S_Funny3)
- Suppose we add the following subtyping rule:
- ------------- (S_Funny4)
- Suppose we add the following evaluation rule:
- ---------------- (ST_Funny5)
- Suppose we add the same evaluation rule and a new typing rule:
- ---------------- (ST_Funny5)
- --------------------- (T_Funny6)
- Suppose we change the arrow subtyping rule to:
S1 <: T1 S2 <: T2
- ---------------------- (S_Arrow') S1->S2 <: T1->T2
Exercise: Adding Products
Exercise: 4 stars, optional (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 well-formedness relation in the obvious way.
- Extend the operational semantics with the same reduction rules
as in the last chapter.
- Extend the subtyping relation with this rule:
S1 <: T1 S2 <: T2
- -------------------- (Sub_Prod) S1 * S2 <: T1 * T2
- Extend the typing relation with the same rules for pairs and
projections as in the last chapter.
- 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.) ☐