SubSubtyping
A Motivating Example
Person = {name:String, age:Nat} Student = {name:String, age:Nat, gpa:Nat}
(λr:Person. (r.age)+1) {name="Pat",age=21,gpa=1}This is a shame.
- 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
- Invoking a method m of an object o on some arguments a1..an consists of projecting out the m field of o and applying it to a1..an.
- An object belonging to a subclass must provide all the
methods and fields of one belonging to a superclass, plus
possibly some more.
- Thus a subclass object can be used anywhere a superclass
object is expected.
- Very handy for organizing large libraries
Of course, real OO languages have lots of other features...
- mutable fields
- private
- method inheritance
- static fields
- 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) -> DIs it safe to allow the application g f?
C->Student <: C->PersonI.e., arrow is covariant in its right-hand argument.
Now suppose we have:
f : Person -> C g : (Student -> C) -> DIs it safe to allow the application g f?
Person->C <: Student->CI.e., arrow is contravariant in its left-hand argument.
Putting these together...
T_{1} <: S_{1} S_{2} <: T_{2} | (S_Arrow) |
S_{1}->S_{2} <: T_{1}->T_{2} |
Suppose we have S <: T and U <: V. Which of the following
subtyping assertions is false?
(1) S×U <: T×V
(2) T→U <: S→U
(3) (S→U) → (S×V) <: (S→U) → (T×U)
(4) (S×U) → V <: (T×U) → V
(5) S→U <: S→V
Suppose again that we have S <: T and U <: V. Which of the
following is incorrect?
(1) (T→T)×U <: (S→T)×V
(2) T→U <: S→V
(3) (S→U)->(S→V) <: (T→U)->(T→V)
(4) (S→V)→V <: (T→U)→V
(5) S->(V→U) <: S->(U→U)
Exercise: 2 stars (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.
Records
The basic intuition about subtyping for record types 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:
for each jk in j1..jn, | |
∃ip in i1..im, such that | |
jk=ip and Sp <: Tk | (S_Rcd) |
{i1:S_{1}...im:Sm} <: {j1:T_{1}...jn:Tn} |
First, adding fields to the end of a record type gives a subtype:
n > m | (S_RcdWidth) |
{i1:T_{1}...in:Tn} <: {i1:T_{1}...im:Tm} |
Second, we can apply subtyping inside the components of a compound record type:
S_{1} <: T_{1} ... Sn <: Tn | (S_RcdDepth) |
{i1:S_{1}...in:Sn} <: {i1:T_{1}...in:Tn} |
Third, we need to be able to reorder fields. For example, we might expect that {name:String, gpa:Nat, age:Nat} <: Person. 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 need:
{i1:S_{1}...in:Sn} is a permutation of {i1:T_{1}...in:Tn} | (S_RcdPerm) |
{i1:S_{1}...in:Sn} <: {i1:T_{1}...in:Tn} |
It is worth noting that full-blown language designs may choose not to adopt all of these subtyping rules. For example, in Java:
- 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).
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) {i1:T_{1}...in:Tn} <: {i1:T_{1}...im:Tm} S_{1} <: T_{1} ... Sn <: Tn (S_RcdDepth) {i1:S_{1}...in:Sn} <: {i1:T_{1}...in:Tn} {i1:S_{1}...in:Sn} is a permutation of {i1:T_{1}...in:Tn} (S_RcdPerm) {i1:S_{1}...in:Sn} <: {i1:T_{1}...in:Tn}
Suppose we have S <: T and U <: V. Which of the following
subtyping assertions is false?
(1) S×U <: Top
(2) {i1:S,i2:T}→U <: {i1:S,i2:T,i3:V}→U
(3) (S→T) → (Top → Top) <: (S→T) → Top
(4) (Top → Top) → V <: Top → V
(5) S->{i1:U,i2:V} <: S->{i2:V,i1:U}
How about these?
(1) {i1:Top} <: Top
(2) Top → (Top → Top) <: Top → Top
(3) {i1:T} → {i1:T} <: {i1:T,i2:S} → Top
(4) {i1:T,i2:V,i3:V} <: {i1:S,i2:U}*{i3:V}
(5) Top->{i1:U,i2:V} <: {i1:S}->{i2:V,i1:V}
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_{1},
(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_{2} 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_{1},
(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_{2} 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
S0, 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 S0, 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 ⊢ (λp:T×Top. p.fst) ((λz: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 ⊢ (λp:(A→A × B→B). p) ((λz:A.z), (λz: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 ⊢ (λp:(A×T). (p.snd) (p.fst)) (a , λz: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 ⊢ (λp:(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 ⊢ (λx:T. x x) t : S
☐
empty ⊢ (λx:T. x x) t : S
Exercise: 2 stars (smallest_2)
What is the smallest type T that makes the following assertion true?
empty ⊢ (λx:Top. x) ((λz:A.z) , (λz:B.z)) : T
☐
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 t_{1} ⇒
tabs y T (if eq_id_dec 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}).
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.
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.
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"
].
Definition context := id → (option ty).
Definition empty : context := (fun _ ⇒ None).
Definition extend (Γ : context) (x:id) (T : ty) :=
fun x' ⇒ if eq_id_dec x x' then Some T else Γ x'.
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},
(extend 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.
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" ].
Properties
- Statements of these theorems don't need to change, compared
to pure STLC
- But proofs are a bit more involved, to account for the additional flexibility in the typing relation
Inversion Lemmas for Subtyping
- Bool is the only subtype of Bool
- every subtype of an arrow type is an arrow type.
Lemma sub_inversion_Bool : ∀U,
U <: TBool →
U = TBool.
Proof with auto.
intros U Hs.
remember TBool as V.
(* FILL IN HERE *) Admitted.
intros U Hs.
remember TBool as V.
(* FILL IN HERE *) 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.
(* FILL IN HERE *) Admitted.
intros U V1 V2 Hs.
remember (TArrow V1 V2) as V.
generalize dependent V2. generalize dependent V1.
(* FILL IN HERE *) Admitted.
Canonical Forms
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.
(* 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.
has_type_cases (induction Hty) Case; try solve by inversion...
Case "T_Sub".
subst. apply sub_inversion_Bool in H. subst...
Qed.
intros Γ 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 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 λx:S_{1}.s_{2} for some x, S_{1}, and s_{2}. But then
(λx:S_{1}.s_{2}) 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.
has_type_cases (induction Ht) Case;
intros HeqGamma; subst...
Case "T_Var".
inversion H.
Case "T_App".
right.
destruct IHHt1; subst...
SCase "t_{1} is a value".
destruct IHHt2; subst...
SSCase "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})...
SSCase "t_{2} steps".
inversion H0 as [t_{2}' Hstp]. ∃(tapp t_{1} t_{2}')...
SCase "t_{1} steps".
inversion H as [t_{1}' Hstp]. ∃(tapp t_{1}' t_{2})...
Case "T_If".
right.
destruct IHHt1.
SCase "t_{1} is a value"...
assert (t_{1} = ttrue ∨ t_{1} = tfalse)
by (eapply canonical_forms_of_Bool; eauto).
inversion H0; subst...
inversion H. rename x into t_{1}'. eauto.
Qed.
intros t T Ht.
remember empty as Γ.
revert HeqGamma.
has_type_cases (induction Ht) Case;
intros HeqGamma; subst...
Case "T_Var".
inversion H.
Case "T_App".
right.
destruct IHHt1; subst...
SCase "t_{1} is a value".
destruct IHHt2; subst...
SSCase "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})...
SSCase "t_{2} steps".
inversion H0 as [t_{2}' Hstp]. ∃(tapp t_{1} t_{2}')...
SCase "t_{1} steps".
inversion H as [t_{1}' Hstp]. ∃(tapp t_{1}' t_{2})...
Case "T_If".
right.
destruct IHHt1.
SCase "t_{1} is a value"...
assert (t_{1} = ttrue ∨ t_{1} = tfalse)
by (eapply canonical_forms_of_Bool; eauto).
inversion H0; 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 ⊢ λx:S_{1}.t_{2} : 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
∧ (extend 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.
has_type_cases (induction H) Case;
inversion Heqt; subst; intros; try solve by inversion.
Case "T_Abs".
∃T_{12}...
Case "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.
has_type_cases (induction H) Case;
inversion Heqt; subst; intros; try solve by inversion.
Case "T_Abs".
∃T_{12}...
Case "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.
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.
intros Γ 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 : ∀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.
has_type_cases (induction Hty) Case; intros;
inversion Heqt; subst; try solve by inversion.
Case "T_App".
∃T_{1}...
Case "T_Sub".
destruct IHHty as [U1 [Hty1 Hty2]]...
Qed.
intros Γ t_{1} t_{2} T_{2} Hty.
remember (tapp t_{1} t_{2}) as t.
has_type_cases (induction Hty) Case; intros;
inversion Heqt; subst; try solve by inversion.
Case "T_App".
∃T_{1}...
Case "T_Sub".
destruct IHHty as [U1 [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.
has_type_cases (induction Htyp) Case;
inversion Heqtu; subst; intros...
Qed.
intros Γ T Htyp. remember ttrue as tu.
has_type_cases (induction Htyp) Case;
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.
has_type_cases (induction Htyp) Case;
inversion Heqtu; subst; intros...
Qed.
intros Γ T Htyp. remember tfalse as tu.
has_type_cases (induction Htyp) Case;
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.
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.
intros Γ t_{1} t_{2} t_{3} T Hty.
remember (tif t_{1} t_{2} t_{3}) 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 : ∀Γ T,
Γ ⊢ tunit ∈ T →
TUnit <: T.
Proof with eauto.
intros Γ T Htyp. remember tunit as tu.
has_type_cases (induction Htyp) Case;
inversion Heqtu; subst; intros...
Qed.
intros Γ 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 S_{1} s_{2} T_{1} T_{2},
empty ⊢ (tabs x S_{1} s_{2}) ∈ (TArrow T_{1} T_{2}) →
T_{1} <: S_{1}
∧ (extend 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 [U1 [U2 [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 [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 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_if1 : ∀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_if2 : ∀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_if3 : ∀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'.
has_type_cases (induction H) Case;
intros Γ' 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 T_{1}...
Case "T_If".
apply T_If...
Qed.
intros. generalize dependent Γ'.
has_type_cases (induction H) Case;
intros Γ' 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 T_{1}...
Case "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.
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.
intros x t T Γ 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.
Lemma substitution_preserves_typing : ∀Γ x U v t S,
(extend 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.
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 [T_{1} [Htypt1 Htypt2]].
eapply T_App...
Case "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 (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 Γ x U) ⊢ t_{1} ∈ TBool
∧ (extend Γ x U) ⊢ t_{2} ∈ S
∧ (extend Γ x U) ⊢ t_{3} ∈ 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.
The proof of preservation now proceeds pretty much as in earlier
chapters, using the substitution lemma at the appropriate point
and again using inversion lemmas 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 : T and t ⇒ t', then empty ⊢ t' :
T.
Proof: Let t and T be given such that empty ⊢ t : T. We
go 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.
Theorem preservation : ∀t t' T,
empty ⊢ t ∈ T →
t ⇒ t' →
empty ⊢ t' ∈ T.
intros Γ x U v t S Htypt Htypv.
generalize dependent S. generalize dependent Γ.
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 [T_{1} [Htypt1 Htypt2]].
eapply T_App...
Case "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 (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 Γ x U) ⊢ t_{1} ∈ TBool
∧ (extend Γ x U) ⊢ t_{2} ∈ S
∧ (extend Γ x U) ⊢ t_{3} ∈ 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 t_{1} t_{2} and types T_{1} 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'.
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.
intros t t' T HT.
remember empty as Γ. 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.