# StlcThe Simply Typed Lambda-Calculus

(* \$Date: 2012-07-25 16:43:16 -0400 (Wed, 25 Jul 2012) \$ *)

Require Export Types.

# The Simply Typed Lambda-Calculus

The simply typed lambda-calculus (STLC) is a tiny core calculus embodying the key concept of functional abstraction, which shows up in pretty much every real-world programming language in some form (functions, procedures, methods, etc.).
We will follow exactly the same pattern as in the previous chapter when formalizing this calculus (syntax, small-step semantics, typing rules) and its main properties (progress and preservation). The new technical challenges (which will take some work to deal with) all arise from the mechanisms of variable binding and substitution.

## Overview

The STLC is built on some collection of base types — booleans, numbers, strings, etc. The exact choice of base types doesn't matter — the construction of the language and its theoretical properties work out pretty much the same — so for the sake of brevity let's take just Bool for the moment. At the end of the chapter we'll see how to add more base types, and in later chapters we'll enrich the pure STLC with other useful constructs like pairs, records, subtyping, and mutable state.
Starting from the booleans, we add three things:
• variables
• function abstractions
• application
This gives us the following collection of abstract syntax constructors (written out here in informal BNF notation — we'll formalize it below).
Informal concrete syntax:
```       t ::= x                       variable
| \x:T.t1                 abstraction
| t1 t2                   application
| true                    constant true
| false                   constant false
| if t1 then t2 else t3   conditional
```
The \ symbol in a function abstraction \x:T.t1 is often written as a greek "lambda" (hence the name of the calculus). The variable x is called the parameter to the function; the term t1 is its body. The annotation :T specifies the type of arguments that the function can be applied to.
Some examples:
• \x:Bool. x
The identity function for booleans.
• (\x:Bool. x) true
The identity function for booleans, applied to the boolean true.
• \x:Bool. if x then false else true
The boolean "not" function.
• \x:Bool. true
The constant function that takes every (boolean) argument to true.
• \x:Bool. \y:Bool. x
A two-argument function that takes two booleans and returns the first one. (Note that, as in Coq, a two-argument function is really a one-argument function whose body is also a one-argument function.)
• (\x:Bool. \y:Bool. x) false true
A two-argument function that takes two booleans and returns the first one, applied to the booleans false and true. Note that, as in Coq, application associates to the left — i.e., this expression is parsed as ((\x:Bool. \y:Bool. x) false) true.
• \f:BoolBool. f (f true)
A higher-order function that takes a function f (from booleans to booleans) as an argument, applies f to true, and applies f again to the result.
• (\f:BoolBool. f (f true)) (\x:Bool. false)
The same higher-order function, applied to the constantly false function.
As the last several examples show, the STLC is a language of higher-order functions: we can write down functions that take other functions as arguments and/or return other functions as results.
Another point to note is that the STLC doesn't provide any primitive syntax for defining named functions — all functions are "anonymous." We'll see in chapter MoreStlc that it is easy to add named functions to what we've got — indeed, the fundamental naming and binding mechanisms are exactly the same.
The types of the STLC include Bool, which classifies the boolean constants true and false as well as more complex computations that yield booleans, plus arrow types that classify functions.
```      T ::= Bool
| T1 -> T2
```
For example:
• \x:Bool. false has type BoolBool
• \x:Bool. x has type BoolBool
• (\x:Bool. x) true has type Bool
• \x:Bool. \y:Bool. x has type BoolBoolBool (i.e. Bool (BoolBool))
• (\x:Bool. \y:Bool. x) false has type BoolBool
• (\x:Bool. \y:Bool. x) false true has type Bool
• \f:BoolBool. f (f true) has type (BoolBool) Bool
• (\f:BoolBool. f (f true)) (\x:Bool. false) has type Bool

## Syntax

Module STLC.

### Types

Inductive ty : Type :=
| TBool : ty
| TArrow : ty ty ty.

### Terms

Inductive tm : Type :=
| tvar : id tm
| tapp : tm tm tm
| tabs : id ty tm tm
| ttrue : tm
| tfalse : tm
| tif : tm tm tm 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" ].

Note that an abstraction \x:T.t (formally, tabs x T t) is always annotated with the type T of its parameter, in contrast to Coq (and other functional languages like ML, Haskell, etc.), which use type inference to fill in missing annotations. We're not considering type inference here, to keep things simple.
Some examples...

Definition x := (Id 0).
Definition y := (Id 1).
Definition z := (Id 2).
Hint Unfold x.
Hint Unfold y.
Hint Unfold z.

idB = \x:Bool. x

Notation idB :=
(tabs x TBool (tvar x)).

idBB = \x:BoolBool. x

Notation idBB :=
(tabs x (TArrow TBool TBool) (tvar x)).

idBBBB = \x:(BoolBool)->(BoolBool). x

Notation idBBBB :=
(tabs x (TArrow (TArrow TBool TBool)
(TArrow TBool TBool))
(tvar x)).

k = \x:Bool. \y:Bool. x

Notation k := (tabs x TBool (tabs y TBool (tvar x))).

(We write these as Notations rather than Definitions to make things easier for auto.)

## Operational Semantics

To define the small-step semantics of STLC terms, we begin — as always — by defining the set of values. Next, we define the critical notions of free variables and substitution, which are used in the reduction rule for application expressions. And finally we give the small-step relation itself.

### Values

To define the values of the STLC, we have a few cases to consider.
First, for the boolean part of the language, the situation is clear: true and false are the only values. (An if expression is never a value.)
Second, an application is clearly not a value: It represents a function being invoked on some argument, which clearly still has work left to do.
Third, for abstractions, we have a choice:
• We can say that \x:T.t1 is a value only when t1 is a value — i.e., only if the function's body has been reduced (as much as it can be without knowing what argument it is going to be applied to).
• Or we can say that \x:T.t1 is always a value, no matter whether t1 is one or not — in other words, we can say that reduction stops at abstractions.
Coq (in its built-in functional programming langauge) makes the first choice — for example,
Eval simpl in (fun x:bool => 3 + 4)
yields fun x:bool => 7. But most real-world functional programming languages make the second choice — reduction of a function's body only begins when the function is actually applied to an argument. We also make the second choice here.
Finally, having made the choice not to reduce under abstractions, we don't need to worry about whether variables are values, since we'll always be reducing programs "from the outside in," and that means the step relation will always be working with closed terms (ones with no free variables).

Inductive value : tm Prop :=
| v_abs : x T t,
value (tabs x T t)
| t_true :
value ttrue
| t_false :
value tfalse.

Hint Constructors value.

### Free Variables and Substitution

Now we come to the heart of the matter: the operation of substituting one term for a variable in another term.
This operation will be used below to define the operational semantics of function application, where we will need to substitute the argument term for the function parameter in the function's body. For example, we reduce
(\x:Bool. if x then true else xfalse
to false by substituting false for the parameter x in the body of the function. In general, we need to be able to substitute some given term s for occurrences of some variable x in another term t. In informal discussions, this is usually written [x:=s]t and pronounced "substitute x with s in t."
Here are some examples:
• [x:=true] (if x then x else false) yields if true then true else false
• [x:=true] x yields true
• [x:=true] (if x then x else y) yields if true then true else y
• [x:=true] y yields y
• [x:=true] false yields false (vacuous substitution)
• [x:=true] (\y:Bool. if y then x else false) yields \y:Bool. if y then true else false
• [x:=true] (\y:Bool. x) yields \y:Bool. true
• [x:=true] (\y:Bool. y) yields \y:Bool. y
• [x:=true] (\x:Bool. x) yields \x:Bool. x
The last example is very important: substituting x with true in \x:Bool. x does not yield \x:Bool. true! The reason for this is that the x in the body of \x:Bool. x is bound by the abstraction: it is a new, local name that just happens to be spelled the same as some global name x.
Here is the definition, informally...
[x:=s]x = s
[x:=s]y = y                                   if x <> y
[x:=s](\x:T11.t12)   = \x:T11. t12
[x:=s](\y:T11.t12)   = \y:T11. [x:=s]t12      if x <> y
[x:=s](t1 t2)        = ([x:=s]t1) ([x:=s]t2)
[x:=s]true           = true
[x:=s]false          = false
[x:=s](if t1 then t2 else t3) =
if [x:=s]t1 then [x:=s]t2 else [x:=s]t3
... and formally:

Fixpoint subst (x:id) (s:tm) (t:tm) : tm :=
match t with
| tvar x' =>
if beq_id x x' then s else t
| tabs x' T t1 =>
tabs x' T (if beq_id x x' 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)
end.

Notation "'[' x ':=' s ']' t" := (subst x s t) (at level 20).

Technical note: Substitution becomes trickier to define if we consider the case where s, the term being substituted for a variable in some other term, may itself contain free variables. Since we are only interested here in defining the step relation on closed terms (i.e., terms like \x:Bool. x y that mention variables are not bound by some enclosing lambda), we can avoid this extra complexity here.

### Reduction

The small-step reduction relation for STLC now follows the same pattern as the ones we have seen before. Intuitively, to reduce a function application, we first reduce its left-hand side until it becomes a literal function; then we reduce its right-hand side (the argument) until it is also a value; and finally we substitute the argument for the bound variable in the body of the function. This last rule, written informally as
(\x:T.t12v2  [x:=v2]t12
is traditionally called "beta-reduction".
 value v2 (ST_AppAbs) (\x:T.t12) v2 ⇒ [x:=v2]t12
 t1 ⇒ t1' (ST_App1) t1 t2 ⇒ t1' t2
 value v1 t2 ⇒ t2' (ST_App2) v1 t2 ⇒ v1 t2'
... plus the usual rules for booleans:
 (ST_IfTrue) (if true then t1 else t2) ⇒ t1
 (ST_IfFalse) (if false then t1 else t2) ⇒ t2
 t1 ⇒ t1' (ST_If) (if t1 then t2 else t3) ⇒ (if t1' then t2 else t3)

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.

Notation multistep := (multi step).
Notation "t1 '⇒*' t2" := (multistep t1 t2) (at level 40).

### Examples

Example:
((\x:BoolBool. x) (\x:Bool. x)) ⇒* (\x:Bool. x)
i.e.
(idBB idB⇒* idB

Lemma step_example1 :
(tapp idBB idB) ⇒* idB.
Proof.
eapply multi_step.
apply ST_AppAbs.
apply v_abs.
simpl.
apply multi_refl. Qed.

A more automatic proof

Lemma step_example1' :
(tapp idBB idB) ⇒* idB.
Proof. normalize. Qed.

Lemma step_example2 :
(tapp idBB (tapp idBB idB)) ⇒* idB.
Proof.
eapply multi_step.
apply ST_App2. auto.
apply ST_AppAbs. auto.
eapply multi_step.
apply ST_AppAbs. simpl. auto.
simpl. apply multi_refl. Qed.

Again, we can use the normalize tactic from above to simplify the proof.

Lemma step_example2' :
(tapp idBB (tapp idBB idB)) ⇒* idB.
Proof.
normalize.
Qed.

#### Exercise: 2 stars (step_example3)

Try to do this one both with and without normalize.

Lemma step_example3 :
(tapp (tapp idBBBB idBB) idB)
⇒* idB.
Proof.
(* FILL IN HERE *) Admitted.

(* FILL IN HERE *)

## Typing

### Contexts

Question: What is the type of the term "x y"?
Answer: It depends on the types of x and y!
I.e., in order to assign a type to a term, we need to know what assumptions we should make about the types of its free variables.
This leads us to a three-place "typing judgment", informally written Γ t : T, where Γ is a "typing context" — a mapping from variables to their types.
We hide the definition of partial maps in a module since it is actually defined in SfLib.

Module PartialMap.

Definition partial_map (A:Type) := id option A.

Definition empty {A:Type} : partial_map A := (fun _ => None).

Informally, we'll write Γ, x:T for "extend the partial function Γ to also map x to T." Formally, we use the function extend to add a binding to a partial map.

Definition extend {A:Type} (Γ : partial_map A) (x:id) (T : A) :=
fun x' => if beq_id x x' then Some T else Γ x'.

Lemma extend_eq : A (ctxt: partial_map A) x T,
(extend ctxt x T) x = Some T.
Proof.
intros. unfold extend. rewrite beq_id_refl. auto.
Qed.

Lemma extend_neq : A (ctxt: partial_map A) x1 T x2,
beq_id x2 x1 = false
(extend ctxt x2 T) x1 = ctxt x1.
Proof.
intros. unfold extend. rewrite H. auto.
Qed.

End PartialMap.

Definition context := partial_map ty.

### Typing Relation

 Γ x = T (T_Var) Γ ⊢ x : T
 Γ , x:T11 ⊢ t12 : T12 (T_Abs) Γ ⊢ \x:T11.t12 : T11->T12
 Γ ⊢ t1 : T11->T12 Γ ⊢ t2 : T11 (T_App) Γ ⊢ t1 t2 : T12
 (T_True) Γ ⊢ true : Bool
 (T_False) Γ ⊢ false : Bool
 Γ ⊢ t1 : Bool    Γ ⊢ t2 : T    Γ ⊢ t3 : T (T_If) Γ ⊢ if t1 then t2 else t3 : T

Inductive has_type : context tm ty Prop :=
| T_Var : Γ x T,
Γ x = Some T
has_type Γ (tvar x) T
| T_Abs : Γ x T11 T12 t12,
has_type (extend Γ x T11) t12 T12
has_type Γ (tabs x T11 t12) (TArrow T11 T12)
| T_App : T11 T12 Γ t1 t2,
has_type Γ t1 (TArrow T11 T12)
has_type Γ t2 T11
has_type Γ (tapp t1 t2) T12
| T_True : Γ,
has_type Γ ttrue TBool
| T_False : Γ,
has_type Γ tfalse TBool
| T_If : t1 t2 t3 T Γ,
has_type Γ t1 TBool
has_type Γ t2 T
has_type Γ t3 T
has_type Γ (tif t1 t2 t3) T.

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" ].

Hint Constructors has_type.

### Examples

Example typing_example_1 :
has_type empty (tabs x TBool (tvar x)) (TArrow TBool TBool).
Proof.
apply T_Abs. apply T_Var. reflexivity. Qed.

Note that since we added the has_type constructors to the hints database, auto can actually solve this one immediately.

Example typing_example_1' :
has_type empty (tabs x TBool (tvar x)) (TArrow TBool TBool).
Proof. auto. Qed.

Hint Unfold beq_id beq_nat extend.

Another example:
empty  \x:A. \y:AA. y (y x))
: A  (AA A.

Example typing_example_2 :
has_type empty
(tabs x TBool
(tabs y (TArrow TBool TBool)
(tapp (tvar y) (tapp (tvar y) (tvar x)))))
(TArrow TBool (TArrow (TArrow TBool TBool) TBool)).
Proof with auto using extend_eq.
apply T_Abs.
apply T_Abs.
eapply T_App. apply T_Var...
eapply T_App. apply T_Var...
apply T_Var...
Qed.

#### Exercise: 2 stars, optional (typing_example_2_full)

Prove the same result without using auto, eauto, or eapply.

Example typing_example_2_full :
has_type empty
(tabs x TBool
(tabs y (TArrow TBool TBool)
(tapp (tvar y) (tapp (tvar y) (tvar x)))))
(TArrow TBool (TArrow (TArrow TBool TBool) TBool)).
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 2 stars (typing_example_3)

Formally prove the following typing derivation holds:
empty  (\x:BoolB. \y:BoolBool. \z:Bool.
y (x z))
: T.

Example typing_example_3 :
T,
has_type empty
(tabs x (TArrow TBool TBool)
(tabs y (TArrow TBool TBool)
(tabs z TBool
(tapp (tvar y) (tapp (tvar x) (tvar z))))))
T.

Proof with auto.
(* FILL IN HERE *) Admitted.
We can also show that terms are not typable. For example, let's formally check that there is no typing derivation assigning a type to the term \x:Bool. \y:Bool, x y — i.e.,
~  T,
empty  (\x:Bool. \y:Boolx y) : T.

Example typing_nonexample_1 :
~ T,
has_type empty
(tabs x TBool
(tabs y TBool
(tapp (tvar x) (tvar y))))
T.
Proof.
intros Hc. inversion Hc.
(* The clear tactic is useful here for tidying away bits of
the context that we're not going to need again. *)

inversion H. subst. clear H.
inversion H5. subst. clear H5.
inversion H4. subst. clear H4.
inversion H2. subst. clear H2.
inversion H5. subst. clear H5.
(* rewrite extend_neq in H1. rewrite extend_eq in H1. *)
inversion H1. Qed.

#### Exercise: 3 stars (typing_nonexample_3)

Another nonexample:
~ ( S T,
empty  (\x:S. x x) : T).

Example typing_nonexample_3 :
~ (S, T,
has_type empty
(tabs x S
(tapp (tvar x) (tvar x)))
T).
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 1 star, optional (typing_statements)

Which of the following propositions are provable?
• y:Bool \x:Bool.x : BoolBool
• T, empty (\y:BoolBool. \x:Bool. y x) : T
• T, empty (\y:BoolBool. \x:Bool. x y) : T
• S, x:S (\y:BoolBool. y) x : S
• S, T, x:S (x x x) : T

#### Exercise: 1 star (more_typing_statements)

Which of the following propositions are provable? For the ones that are, give witnesses for the existentially bound variables.
• T, empty (\y:BBB. \x:B, y x) : T
• T, empty (\x:AB, \y:BC, \z:A, y (x z)):T
• S, U, T, x:S, y:U \z:A. x (y z) : T
• S, T, x:S \y:A. x (x y) : T
• S, U, T, x:S x (\z:U. z x) : T

## Properties

### Progress

The progress theorem tells us that closed, well-typed terms are not stuck: either a well-typed term is a value, or it can take an evaluation step.

Theorem progress : t T,
has_type empty t T
value t t', t t'.

Proof: by induction on the derivation of t : T.
• The last rule of the derivation cannot be T_Var, since a variable is never well typed in an empty context.
• The T_True, T_False, and T_Abs cases are trivial, since in each of these cases we know immediately that t is a value.
• If the last rule of the derivation was T_App, then t = t1 t2, and we know that t1 and t2 are also well typed in the empty context; in particular, there exists a type T2 such that t1 : T2 T and t2 : T2. By the induction hypothesis, either t1 is a value or it can take an evaluation step.
• If t1 is a value, we now consider t2, which by the other induction hypothesis must also either be a value or take an evaluation step.
• Suppose t2 is a value. Since t1 is a value with an arrow type, it must be a lambda abstraction; hence t1 t2 can take a step by ST_AppAbs.
• Otherwise, t2 can take a step, and hence so can t1 t2 by ST_App2.
• If t1 can take a step, then so can t1 t2 by ST_App1.
• If the last rule of the derivation was T_If, then t = if t1 then t2 else t3, where t1 has type Bool. By the IH, t1 is either a value or takes a step.
• If t1 is a value, then since it has type Bool it must be either true or false. If it is true, then t steps to t2; otherwise it steps to t3.
• Otherwise, t1 takes a step, and therefore so does t (by ST_If).

Proof with eauto.
intros t T Ht.
remember (@empty ty) as Γ.
has_type_cases (induction Ht) Case; subst Γ...
Case "T_Var".
(* contradictory: variables cannot be typed in an
empty context *)

inversion H.

Case "T_App".
(* t = t1 t2.  Proceed by cases on whether t1 is a
value or steps... *)

right. destruct IHHt1...
SCase "t1 is a value".
destruct IHHt2...
SSCase "t2 is also a value".
(* Since t1 is a value and has an arrow type, it
must be an abs. Sometimes this is proved separately
and called a "canonical forms" lemma. *)

inversion H; subst. ([x0:=t2]t)...
solve by inversion. solve by inversion.
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".
(* Since t1 is a value of boolean type, it must
be true or false *)

inversion H; subst. solve by inversion.
SSCase "t1 = true". eauto.
SSCase "t1 = false". eauto.

SCase "t1 also steps".
inversion H as [t1' Hstp]. (tif t1' t2 t3)...
Qed.

#### Exercise: 3 stars, optional (progress_from_term_ind)

Show that progress can also be proved by induction on terms instead of induction on typing derivations.

Theorem progress' : t T,
has_type empty t T
value t t', t t'.
Proof.
intros t.
t_cases (induction t) Case; intros T Ht; auto.
(* FILL IN HERE *) Admitted.

### Free Occurrences

A variable x appears free in a term t if t contains some occurrence of x that is not under an abstraction labeled x. For example:
• y appears free, but x does not, in \x:TU. x y
• both x and y appear free in (\x:TU. x y) x
• no variables appear free in \x:TU. \y:T. x y

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).

Tactic Notation "afi_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "afi_var"
| Case_aux c "afi_app1" | Case_aux c "afi_app2"
| Case_aux c "afi_abs"
| Case_aux c "afi_if1" | Case_aux c "afi_if2"
| Case_aux c "afi_if3" ].

Hint Constructors appears_free_in.

A term in which no variables appear free is said to be closed.

Definition closed (t:tm) :=
x, ~ appears_free_in x t.

### Substitution

We first need a technical lemma connecting free variables and typing contexts. If a variable x appears free in a term t, and if we know t is well typed in context Γ, then it must be the case that Γ assigns a type to x.

Lemma free_in_context : x t T Γ,
appears_free_in x t
has_type Γ t T
T', Γ x = Some T'.

Proof: We show, by induction on the proof that x appears free in t, that, for all contexts Γ, if t is well typed under Γ, then Γ assigns some type to x.
• If the last rule used was afi_var, then t = x, and from the assumption that t is well typed under Γ we have immediately that Γ assigns a type to x.
• If the last rule used was afi_app1, then t = t1 t2 and x appears free in t1. Since t is well typed under Γ, we can see from the typing rules that t1 must also be, and the IH then tells us that Γ assigns x a type.
• Almost all the other cases are similar: x appears free in a subterm of t, and since t is well typed under Γ, we know the subterm of t in which x appears is well typed under Γ as well, and the IH gives us exactly the conclusion we want.
• The only remaining case is afi_abs. In this case t = \y:T11.t12, and x appears free in t12; we also know that x is different from y. The difference from the previous cases is that whereas t is well typed under Γ, its body t12 is well typed under (Γ, y:T11), so the IH allows us to conclude that x is assigned some type by the extended context (Γ, y:T11). To conclude that Γ assigns a type to x, we appeal to lemma extend_neq, noting that x and y are different variables.

Proof.
intros x t T Γ H H0. generalize dependent Γ.
generalize dependent T.
afi_cases (induction H) Case;
intros; try solve [inversion H0; eauto].
Case "afi_abs".
inversion H1; subst.
apply IHappears_free_in in H7.
apply not_eq_beq_id_false in H.
rewrite extend_neq in H7; assumption.
Qed.

Next, we'll need the fact that any term t which is well typed in the empty context is closed — that is, it has no free variables.

#### Exercise: 2 stars (typable_empty__closed)

Corollary typable_empty__closed : t T,
has_type empty t T
closed t.
Proof.
(* FILL IN HERE *) Admitted.
Sometimes, when we have a proof Γ t : T, we will need to replace Γ by a different context Γ'. When is it safe to do this? Intuitively, it must at least be the case that Γ' assigns the same types as Γ to all the variables that appear free in t. In fact, this is the only condition that is needed.

Lemma context_invariance : Γ Γ' t T,
has_type Γ t T
(x, appears_free_in x t Γ x = Γ' x)
has_type Γ' t T.

Proof: By induction on the derivation of Γ t : T.
• If the last rule in the derivation was T_Var, then t = x and Γ x = T. By assumption, Γ' x = T as well, and hence Γ' t : T by T_Var.
• If the last rule was T_Abs, then t = \y:T11. t12, with T = T11 T12 and Γ, y:T11 t12 : T12. The induction hypothesis is that for any context Γ'', if Γ, y:T11 and Γ'' assign the same types to all the free variables in t12, then t12 has type T12 under Γ''. Let Γ' be a context which agrees with Γ on the free variables in t; we must show Γ' \y:T11. t12 : T11 T12.
By T_Abs, it suffices to show that Γ', y:T11 t12 : T12. By the IH (setting Γ'' = Γ', y:T11), it suffices to show that Γ, y:T11 and Γ', y:T11 agree on all the variables that appear free in t12.
Any variable occurring free in t12 must either be y, or some other variable. Γ, y:T11 and Γ', y:T11 clearly agree on y. Otherwise, we note that any variable other than y which occurs free in t12 also occurs free in t = \y:T11. t12, and by assumption Γ and Γ' agree on all such variables, and hence so do Γ, y:T11 and Γ', y:T11.
• If the last rule was T_App, then t = t1 t2, with Γ t1 : T2 T and Γ t2 : T2. One induction hypothesis states that for all contexts Γ', if Γ' agrees with Γ on the free variables in t1, then t1 has type T2 T under Γ'; there is a similar IH for t2. We must show that t1 t2 also has type T under Γ', given the assumption that Γ' agrees with Γ on all the free variables in t1 t2. By T_App, it suffices to show that t1 and t2 each have the same type under Γ' as under Γ. However, we note that all free variables in t1 are also free in t1 t2, and similarly for free variables in t2; hence the desired result follows by the two IHs.

Proof with eauto.
intros.
generalize dependent Γ'.
has_type_cases (induction H) Case; intros; auto.
Case "T_Var".
apply T_Var. rewrite H0...
Case "T_Abs".
apply T_Abs.
apply IHhas_type. intros x1 Hafi.
(* the only tricky step... the Γ' we use to
instantiate is extend Γ x T11 *)

unfold extend. remember (beq_id x0 x1) as e. destruct e...
Case "T_App".
apply T_App with T11...
Qed.

Now we come to the conceptual heart of the proof that reduction preserves types — namely, the observation that substitution preserves types.
Formally, the so-called Substitution Lemma says this: suppose we have a term t with a free variable x, and suppose we've been able to assign a type T to t under the assumption that x has some type U. Also, suppose that we have some other term v and that we've shown that v has type U. Then, since v satisfies the assumption we made about x when typing t, we should be able to substitute v for each of the occurrences of x in t and obtain a new term that still has type T.
Lemma: If Γ,x:U t : T and v : U, then Γ [x:=v]t : T.

Lemma substitution_preserves_typing : Γ x U t t' T,
has_type (extend Γ x U) t T
has_type empty t' U
has_type Γ ([x:=t']t) T.

One technical subtlety in the statement of the lemma is that we assign t' the type U in the empty context — in other words, we assume t' is closed. This assumption considerably simplifies the T_Abs case of the proof (compared to assuming Γ t' : U, which would be the other reasonable assumption at this point) because the context invariance lemma then tells us that t' has type U in any context at all — we don't have to worry about free variables in t' clashing with the variable being introduced into the context by T_Abs.
Proof: We prove, by induction on t, that, for all T and Γ, if Γ,x:U t : T and t' : U, then Γ [x:=t']t : T.
• If t is a variable, there are two cases to consider, depending on whether t is x or some other variable.
• If t = x, then from the fact that Γ, x:U x : T we conclude that U = T. We must show that [x:=t']x = t' has type T under Γ, given the assumption that t' has type U = T under the empty context. This follows from context invariance: if a closed term has type T in the empty context, it has that type in any context.
• If t is some variable y that is not equal to x, then we need only note that y has the same type under Γ, x:U as under Γ.
• If t is an abstraction \y:T11. t12, then the IH tells us, for all Γ' and T', that if Γ',x:U t12 : T' and t' : U, then Γ' [x:=t']t12 : T'.
The substitution in the conclusion behaves differently, depending on whether x and y are the same variable name.
First, suppose x = y. Then, by the definition of substitution, [x:=t']t = t, so we just need to show Γ t : T. But we know Γ,x:U t : T, and since the variable y does not appear free in \y:T11. t12, the context invariance lemma yields Γ t : T.
Second, suppose x <> y. We know Γ,x:U,y:T11 t12 : T12 by inversion of the typing relation, and Γ,y:T11,x:U t12 : T12 follows from this by the context invariance lemma, so the IH applies, giving us Γ,y:T11 [x:=t']t12 : T12. By T_Abs, Γ \y:T11. [x:=t']t12 : T11T12, and by the definition of substitution (noting that x <> y), Γ \y:T11. [x:=t']t12 : T11T12 as required.
• If t is an application t1 t2, the result follows straightforwardly from the definition of substitution and the induction hypotheses.
• The remaining cases are similar to the application case.
Another technical note: This proof is a rare case where an induction on terms, rather than typing derivations, yields a simpler argument. The reason for this is that the assumption has_type (extend Γ x U) t T is not completely generic, in the sense that one of the "slots" in the typing relation — namely the context — is not just a variable, and this means that Coq's native induction tactic does not give us the induction hypothesis that we want. It is possible to work around this, but the needed generalization is a little tricky. The term t, on the other hand, is completely generic.

Proof with eauto.
intros Γ x U t t' T Ht Ht'.
generalize dependent Γ. generalize dependent T.
t_cases (induction t) Case; intros T Γ H;
(* in each case, we'll want to get at the derivation of H *)
inversion H; subst; simpl...
Case "tvar".
rename i into y. remember (beq_id x y) as e. destruct e.
SCase "x=y".
apply beq_id_eq in Heqe. subst.
rewrite extend_eq in H2.
inversion H2; subst. clear H2.
eapply context_invariance... intros x Hcontra.
destruct (free_in_context _ _ T empty Hcontra) as [T' HT']...
inversion HT'.
SCase "x<>y".
apply T_Var. rewrite extend_neq in H2...
Case "tabs".
rename i into y. apply T_Abs.
remember (beq_id x y) as e. destruct e.
SCase "x=y".
eapply context_invariance...
apply beq_id_eq in Heqe. subst.
intros x Hafi. unfold extend.
destruct (beq_id y x)...
SCase "x<>y".
apply IHt. eapply context_invariance...
intros z Hafi. unfold extend.
remember (beq_id y z) as e0. destruct e0...
apply beq_id_eq in Heqe0. subst.
rewrite Heqe...
Qed.

The substitution lemma can be viewed as a kind of "commutation" property. Intuitively, it says that substitution and typing can be done in either order: we can either assign types to the terms t and v separately (under suitable contexts) and then combine them using substitution, or we can substitute first and then assign a type to [x:=v] t — the result is the same either way.

### Preservation

We now have the tools we need to prove preservation: if a closed term t has type T, and takes an evaluation step to t', then t' is also a closed term with type T. In other words, the small-step evaluation relation preserves types.

Theorem preservation : t t' T,
has_type empty t T
t t'
has_type empty t' T.

Proof: by induction on the derivation of t : T.
• We can immediately rule out T_Var, T_Abs, T_True, and T_False as the final rules in the derivation, since in each of these cases t cannot take a step.
• If the last rule in the derivation was T_App, then t = t1 t2. There are three cases to consider, one for each rule that could have been used to show that t1 t2 takes a step to t'.
• If t1 t2 takes a step by ST_App1, with t1 stepping to t1', then by the IH t1' has the same type as t1, and hence t1' t2 has the same type as t1 t2.
• The ST_App2 case is similar.
• If t1 t2 takes a step by ST_AppAbs, then t1 = \x:T11.t12 and t1 t2 steps to [x:=t2]t12; the desired result now follows from the fact that substitution preserves types.
• If the last rule in the derivation was T_If, then t = if t1 then t2 else t3, and there are again three cases depending on how t steps.
• If t steps to t2 or t3, the result is immediate, since t2 and t3 have the same type as t.
• Otherwise, t steps by ST_If, and the desired conclusion follows directly from the induction hypothesis.

Proof with eauto.
remember (@empty ty) as Γ.
intros t t' T HT. generalize dependent t'.
has_type_cases (induction HT) Case;
intros t' HE; subst Γ; subst;
try solve [inversion HE; subst; auto].
Case "T_App".
inversion HE; subst...
(* Most of the cases are immediate by induction,
and eauto takes care of them *)

SCase "ST_AppAbs".
apply substitution_preserves_typing with T11...
inversion HT1...
Qed.

#### Exercise: 2 stars, recommended (subject_expansion_stlc)

An exercise in Types.v asked about the subject expansion property for the simple language of arithmetic and boolean expressions. Does this property hold for STLC? That is, is it always the case that, if t t' and has_type t' T, then has_type t T? If so, prove it. If not, give a counter-example.
(* FILL IN HERE *)

### Type Soundness

#### Exercise: 2 stars, optional (type_soundness)

Put progress and preservation together and show that a well-typed term can never reach a stuck state.

Definition stuck (t:tm) : Prop :=
(normal_form step) t ~ value t.

Corollary soundness : t t' T,
has_type empty t T
t ⇒* t'
~(stuck t').
Proof.
intros t t' T Hhas_type Hmulti. unfold stuck.
intros [Hnf Hnot_val]. unfold normal_form in Hnf.
induction Hmulti.
(* FILL IN HERE *) Admitted.

### Uniqueness of Types

#### Exercise: 3 stars (types_unique)

Another pleasant property of the STLC is that types are unique: a given term (in a given context) has at most one type. Formalize this statement and prove it.

(* FILL IN HERE *)

#### Exercise: 1 star (progress_preservation_statement)

Without peeking, write down the progress and preservation theorems for the simply typed lambda-calculus.

#### Exercise: 2 stars (stlc_variation1)

Suppose we add a new term zap with the following reduction rule:
 (ST_Zap) t ⇒ zap
and the following typing rule:
 (T_Zap) Γ ⊢ zap : T
Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample.
• Determinism of step
• Progress
• Preservation

#### Exercise: 2 stars (stlc_variation2)

Suppose instead that we add a new term foo with the following reduction rules:
 (ST_Foo1) (\x:A. x) ⇒ foo
 (ST_Foo2) foo ⇒ true
Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample.
• Determinism of step
• Progress
• Preservation

#### Exercise: 2 stars (stlc_variation3)

Suppose instead that we remove the rule ST_App1 from the step relation. Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample.
• Determinism of step
• Progress
• Preservation

#### Exercise: 2 stars (stlc_variation4)

Suppose instead that we add the following new rule to the reduction relation:
 (ST_FunnyIfTrue) (if true then t1 else t2) ⇒ true
Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample.
• Determinism of step
• Progress
• Preservation

#### Exercise: 2 stars (stlc_variation5)

Suppose instead that we add the following new rule to the typing relation:
 Γ ⊢ t1 : Bool->Bool->Bool Γ ⊢ t2 : Bool (T_FunnyApp) Γ ⊢ t1 t2 : Bool
Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample.
• Determinism of step
• Progress
• Preservation

#### Exercise: 2 stars (stlc_variation6)

Suppose instead that we add the following new rule to the typing relation:
 Γ ⊢ t1 : Bool Γ ⊢ t2 : Bool (T_FunnyApp') Γ ⊢ t1 t2 : Bool
Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample.
• Determinism of step
• Progress
• Preservation

#### Exercise: 2 stars (stlc_variation7)

Suppose we add the following new rule to the typing relation of the STLC:
 (T_FunnyAbs) ⊢ \x:Bool.t : Bool
Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample.
• Determinism of step
• Progress
• Preservation

End STLC.

# Optional Exercise: STLC with Arithmetic

To see how the STLC might function as the core of a real programming language, let's extend it with a concrete base type of numbers and some constants and primitive operators.

Module STLCArith.

## Syntax and Operational Semantics

To types, we add a base type of natural numbers (and remove booleans, for brevity)

Inductive ty : Type :=
| TArrow : ty ty ty
| TNat : ty.

To terms, we add natural number constants, along with successor, predecessor, multiplication, and zero-testing...

Inductive tm : Type :=
| tvar : id tm
| tapp : tm tm tm
| tabs : id ty tm tm
| tnat : nat tm
| tsucc : tm tm
| tpred : tm tm
| tmult : tm tm tm
| tif0 : tm tm tm 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 "tnat"
| Case_aux c "tsucc" | Case_aux c "tpred"
| Case_aux c "tmult" | Case_aux c "tif0" ].

#### Exercise: 4 stars, optional (stlc_arith)

Finish formalizing the definition and properties of the STLC extended with arithmetic. Specifically:
• Copy the whole development of STLC that we went through above (from the definition of values through the Progress theorem), and paste it into the file at this point.
• Extend the definitions of the subst operation and the step relation to include appropriate clauses for the arithmetic operators.
• Extend the proofs of all the properties of the original STLC to deal with the new syntactic forms. Make sure Coq accepts the whole file.

(* FILL IN HERE *)

End STLCArith.