# StlcThe Simply Typed Lambda-Calculus

(* \$Date: 2011-05-07 16:41:12 -0400 (Sat, 07 May 2011) \$ *)

Require Export Smallstep.
Require Import Relations.

Our next topic, a large one, is type systems — static program analyses that classify expressions according to the "shapes" of their results. We'll begin with a typed version of a very simple language with just booleans and numbers, to introduce the basic ideas of types, typing rules, and the fundamental theorems about type systems: type preservation and progress. Then we'll move on to the simply typed lambda-calculus, which lives at the core of every modern functional programming language (including Coq).

# More Automation

Before we start, let's spend a little time learning to use some of Coq's more powerful automation features...

## The auto and eauto Tactics

The auto tactic solves goals that are solvable by any combination of
• intros,
• apply (with a local hypothesis, by default), and
• reflexivity.
The eauto tactic works just like auto, except that it uses eapply instead of apply.
Using auto is always "safe" in the sense that it will never fail and will never change the proof state: either it completely solves the current goal, or it does nothing.
Here is a contrived example:

Lemma auto_example_1 : P Q R S T U : Prop,
(P Q)
(P R)
(T R)
(S T U)
((PQ) (PS))
T
P
U.
Proof. auto. Qed.

When searching for potential proofs of the current goal, auto and eauto consider the hypotheses in the current context together with a hint database of other lemmas and constructors. Some of the lemmas and constructors we've already seen — e.g., conj, or_introl, and or_intror — are installed in this hint database by default.

Lemma auto_example_2 : P Q R : Prop,
Q
(Q R)
P (Q R).
Proof.
auto. Qed.

We can extend the hint database just for the purposes of one application of auto or eauto by writing auto using .... E.g., if conj, or_introl, and or_intror had not already been in the hint database, we could have done this instead:

Lemma auto_example_2a : P Q R : Prop,
Q
(Q R)
P (Q R).
Proof.
auto using conj, or_introl, or_intror. Qed.

Of course, in any given development there will also be some of our own specific constructors and lemmas that are used very often in proofs. We can add these to the global hint database by writing
Hint Resolve l.
at the top level, where l is a top-level lemma theorem or a constructor of an inductively defined proposition (i.e., anything whose type is an implication). As a shorthand, we can write
Hint Constructors c.
to tell Coq to do a Hint Resolve for all of the constructors from the inductive definition of c.
It is also sometimes necessary to add
Hint Unfold d.
where d is a defined symbol, so that auto knows to expand uses of d and enable further possibilities for applying lemmas that it knows about.
Here are some Hints we will find useful.

Hint Constructors refl_step_closure.
Hint Resolve beq_id_eq beq_id_false_not_eq.

Warning: Just as with Coq's other automation facilities, it is easy to overuse auto and eauto and wind up with proofs that are impossible to understand later!
Also, overuse of eauto can make proof scripts very slow. Get in the habit of using auto most of the time and eauto only when necessary.
For much more detailed information about using auto and eauto, see the chapter UseAuto.v.

## The Proofwith Tactic

If you start a proof by saying Proof with (tactic) instead of just Proof, then writing ... instead of . after a tactic in the body of the proof will try to solve all generated subgoals with tactic (and fail if this doesn't work).
One common use of this facility is "Proof with auto" (or eauto). We'll see many examples of this later in the file.

## The solvebyinversion Tactic

Here's another nice automation feature: it often arises that the context contains a contradictory assumption and we want to use inversion on it to solve the goal. We'd like to be able to say to Coq, "find a contradictory assumption and invert it" without giving its name explicitly.
Doing solve by inversion will find a hypothesis that can be inverted to solve the goal, if there is one. The tactics solve by inversion 2 and solve by inversion 3 are slightly fancier versions which will perform two or three inversions in a row, if necessary, to solve the goal.
(These tactics are not actually built into Coq — their definitions are in Sflib.v.)
Caution: Overuse of solve by inversion can lead to slow proof scripts.

## The trysolve Tactic

If t is a tactic, then try solve [t] is a tactic that
• if t solves the goal, behaves just like t, or
• if t cannot completely solve the goal, does nothing.
More generally, try solve [t1 | t2 | ...] will try to solve the goal by using t1, t2, etc. If none of them succeeds in completely solving the goal, then try solve [t1 | t2 | ...] does nothing.
The fact that it does nothing when it doesn't completely succeed in solving the goal means that there is no harm in using try solve T in situations where T might actually make no sense. In particular, if we put it after a ; it will solve as many subgoals as it can and leave the rest for us to solve by other methods. It will not leave any of the subgoals in a partially solved state.

## The f_equal Tactic

This handy tactic replaces a goal of the form f x1 x2 ... xn = f y1 y2 ... yn, where f is some function, with the subgoals x1 = y1, x2 = y2,...,xn = yn. It is useful for avoiding explicit rewriting steps, and often the generated subgoals can be quickly cleared by auto.

## The normalize Tactic

When experimenting with definitions of programming languages in Coq, we often want to see what a particular concrete term steps to — i.e., we want to find proofs for goals of the form t ⇒* t', where t is a completely concrete term and t' is unknown. These proofs are simple but repetitive to do by hand. Consider for example reducing an arithmetic expression using the small-step relation astep defined in the previous chapter:

Definition astep_many st := refl_step_closure (astep st).
Notation " t '/' st 'a*' t' " := (astep_many st t t')
(at level 40, st at level 39).

Example astep_example1 :
(APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state
a* (ANum 15).
Proof.
apply rsc_step with (APlus (ANum 3) (ANum 12)).
apply AS_Plus2.
apply av_num.
apply AS_Mult.
apply rsc_step with (ANum 15).
apply AS_Plus.
apply rsc_refl.
Qed.

We repeatedly applied rsc_step until we got to a normal form. The proofs that the intermediate steps are possible are simple enough that auto, with appropriate hints, can solve them.

Hint Constructors astep aval.
Example astep_example1' :
(APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state
a* (ANum 15).
Proof.
eapply rsc_step. auto. simpl.
eapply rsc_step. auto. simpl.
apply rsc_refl.
Qed.

The following custom Tactic Notation definition captures this pattern. In addition, before each rsc_step we print out the current goal, so that the user can follow how the term is being evaluated.

Tactic Notation "print_goal" := match goal with ?x => idtac x end.
Tactic Notation "normalize" :=
repeat (print_goal; eapply rsc_step ;
[ (eauto 10; fail) | (instantiate; simpl)]);
apply rsc_refl.

Example astep_example1'' :
(APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state
a* (ANum 15).
Proof.
normalize.
(* At this point in the proof script, the Coq response shows
a trace of how the expression evaluated. *)

Qed.

Finally, this is also useful as a simple way to calculate what the normal form of a term is, by proving a goal with an existential variable in it.

Example astep_example1''' : e',
(APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state
a* e'.
Proof.
eapply ex_intro. normalize.
Qed.

# Typed Arithmetic Expressions

To motivate the discussion of type systems, let's begin as usual with an extremely simple toy language. We want it to have the potential for programs "going wrong" because of runtime type errors, so we need something a tiny bit more complex than the language of constants and addition that we used in Smallstep.v: a single kind of data (just numbers) is too simple, but just two kinds (numbers and booleans) already gives us enough material to tell an interesting story.
The language definition is completely routine. The only thing to notice is that we are not using the asnum/aslist trick that we used in ImpList.v to make all the operations total by forcibly coercing the arguments to + (for example) into numbers. Instead, we simply let terms get stuck if they try to use an operator with the wrong kind of operands: the step relation doesn't relate them to anything.

## Syntax

Inductive tm : Type :=
| tm_true : tm
| tm_false : tm
| tm_if : tm tm tm tm
| tm_zero : tm
| tm_succ : tm tm
| tm_pred : tm tm
| tm_iszero : tm tm.

Inductive bvalue : tm Prop :=
| bv_true : bvalue tm_true
| bv_false : bvalue tm_false.

Inductive nvalue : tm Prop :=
| nv_zero : nvalue tm_zero
| nv_succ : t, nvalue t nvalue (tm_succ t).

Definition value (t:tm) := bvalue t nvalue t.

Hint Constructors bvalue nvalue.
Hint Unfold value.

## Small-Step Reduction

Reserved Notation "t1 '' t2" (at level 40).

Inductive step : tm tm Prop :=
| ST_IfTrue : t1 t2,
(tm_if tm_true t1 t2) t1
| ST_IfFalse : t1 t2,
(tm_if tm_false t1 t2) t2
| ST_If : t1 t1' t2 t3,
t1 t1'
(tm_if t1 t2 t3) (tm_if t1' t2 t3)
| ST_Succ : t1 t1',
t1 t1'
(tm_succ t1) (tm_succ t1')
| ST_PredZero :
(tm_pred tm_zero) tm_zero
| ST_PredSucc : t1,
nvalue t1
(tm_pred (tm_succ t1)) t1
| ST_Pred : t1 t1',
t1 t1'
(tm_pred t1) (tm_pred t1')
| ST_IszeroZero :
(tm_iszero tm_zero) tm_true
| ST_IszeroSucc : t1,
nvalue t1
(tm_iszero (tm_succ t1)) tm_false
| ST_Iszero : t1 t1',
t1 t1'
(tm_iszero t1) (tm_iszero t1')

where "t1 '' t2" := (step t1 t2).

Tactic Notation "step_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "ST_IfTrue" | Case_aux c "ST_IfFalse" | Case_aux c "ST_If"
| Case_aux c "ST_Succ" | Case_aux c "ST_PredZero"
| Case_aux c "ST_PredSucc" | Case_aux c "ST_Pred"
| Case_aux c "ST_IszeroZero" | Case_aux c "ST_IszeroSucc"
| Case_aux c "ST_Iszero" ].

Hint Constructors step.

Notice that the step relation doesn't care about whether expressions make global sense — it just checks that the operation in the next reduction step is being applied to the right kinds of operands. For example, the term tm_succ tm_true cannot take a step, but the almost-as-obviously-nonsensical term
tm_succ (tm_if tm_true tm_true tm_true)
can.

## Normal Forms and Values

The first interesting thing about the step relation in this language is that the strong progress theorem from the Smallstep chapter fails! That is, there are terms that are normal forms (they can't take a step) but not values (because we have not included them in our definition of possible "results of evaluation"). Such terms are stuck.

Notation step_normal_form := (normal_form step).

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

Hint Unfold stuck.

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

Example some_tm_is_stuck :
t, stuck t.
Proof.
(* FILL IN HERE *) Admitted.
However, although values and normal forms are not the same in this language, the former set is included in the latter. This is important because it shows we did not accidentally define things so that some value could still take a step.

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

Hint: You will reach a point in this proof where you need to use an induction to reason about a term that is known to be a numeric value. This induction can be performed either over the term itself or over the evidence that it is a numeric value. The proof goes through in either case, but you will find that one way is quite a bit shorter than the other. For the sake of the exercise, try to complete the proof both ways.

Lemma value_is_nf : t,
value t step_normal_form t.
Proof.
(* FILL IN HERE *) Admitted.

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

Using value_is_nf, we can show that the step relation is also deterministic...

Theorem step_deterministic:
partial_function step.
Proof with eauto.
(* FILL IN HERE *) Admitted.

## Typing

The next critical observation about this language is that, although there are stuck terms, they are all "nonsensical", mixing booleans and numbers in a way that we don't even want to have a meaning. We can easily exclude such ill-typed terms by defining a typing relation that relates terms to the types (either numeric or boolean) of their final results.

Inductive ty : Type :=
| ty_Bool : ty
| ty_Nat : ty.

The typing relation:
In informal notation, the typing relation is often written t : T, pronounced "t has type T." The symbol is called a "turnstile". (Below, we're going to see richer typing relations where an additional "context" argument is written to the left of the turnstile. Here, the context is always empty.)
 (T_True) ⊢ true : Bool
 (T_False) ⊢ false : Bool
 ⊢ t1 : Bool    ⊢ t2 : T    ⊢ t3 : T (T_If) ⊢ if t1 then t2 else t3 : T
 (T_Zero) ⊢ 0 : Nat
 ⊢ t1 : Nat (T_Succ) ⊢ succ t1 : Nat
 ⊢ t1 : Nat (T_Pred) ⊢ pred t1 : Nat
 ⊢ t1 : Nat (T_IsZero) ⊢ iszero t1 : Bool

Inductive has_type : tm ty Prop :=
| T_True :
has_type tm_true ty_Bool
| T_False :
has_type tm_false ty_Bool
| T_If : t1 t2 t3 T,
has_type t1 ty_Bool
has_type t2 T
has_type t3 T
has_type (tm_if t1 t2 t3) T
| T_Zero :
has_type tm_zero ty_Nat
| T_Succ : t1,
has_type t1 ty_Nat
has_type (tm_succ t1) ty_Nat
| T_Pred : t1,
has_type t1 ty_Nat
has_type (tm_pred t1) ty_Nat
| T_Iszero : t1,
has_type t1 ty_Nat
has_type (tm_iszero t1) ty_Bool.

Tactic Notation "has_type_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "T_True" | Case_aux c "T_False" | Case_aux c "T_If"
| Case_aux c "T_Zero" | Case_aux c "T_Succ" | Case_aux c "T_Pred"
| Case_aux c "T_Iszero" ].

Hint Constructors has_type.

### Examples

It's important to realize that the typing relation is a conservative (or static) approximation: it does not calculate the type of the normal form of a term.

Example has_type_1 :
has_type (tm_if tm_false tm_zero (tm_succ tm_zero)) ty_Nat.
Proof.
apply T_If.
apply T_False.
apply T_Zero.
apply T_Succ.
apply T_Zero.
Qed.
(Since we've included all the constructors of the typing relation in the hint database, the auto tactic can actually find this proof automatically.)

Example has_type_not :
~ has_type (tm_if tm_false tm_zero tm_true) ty_Bool.
Proof.
intros Contra. solve by inversion 2. Qed.

#### Exercise: 1 star (succ_hastype_nat__hastype_nat)

Example succ_hastype_nat__hastype_nat : t,
has_type (tm_succ t) ty_Nat
has_type t ty_Nat.
Proof.
(* FILL IN HERE *) Admitted.

## Progress

The typing relation enjoys two critical properties. The first is that well-typed normal forms are values (i.e., not stuck).

#### Exercise: 3 stars, recommended (finish_progress_informal)

Complete the following proof:
Theorem: If t : T, then either t is a value or else t t' for some t'.
Proof: By induction on a derivation of t : T.
• If the last rule in the derivation is T_If, then t = if t1 then t2 else t3, with t1 : Bool, t2 : T and t3 : T. By the IH, either t1 is a value or else t1 can step to some t1'.
• If t1 is a value, then it is either an nvalue or a bvalue. But it cannot be an nvalue, because we know t1 : Bool and there are no rules assigning type Bool to any term that could be an nvalue. So t1 is a bvalue — i.e., it is either true or false. If t1 = true, then t steps to t2 by ST_IfTrue, while if t1 = false, then t steps to t3 by ST_IfFalse. Either way, t can step, which is what we wanted to show.
• If t1 itself can take a step, then, by ST_If, so can t.
(* FILL IN HERE *)

#### Exercise: 3 stars (finish_progress)

Theorem progress : t T,
has_type t T
value t t', t t'.
Proof with auto.
intros t T HT.
has_type_cases (induction HT) Case...
(* The cases that were obviously values, like T_True and
T_False, were eliminated immediately by auto *)

Case "T_If".
right. destruct IHHT1.
SCase "t1 is a value". destruct H.
SSCase "t1 is a bvalue". destruct H.
SSSCase "t1 is tm_true".
t2...
SSSCase "t1 is tm_false".
t3...
SSCase "t1 is an nvalue".
solve by inversion 2. (* on H and HT1 *)
SCase "t1 can take a step".
destruct H as [t1' H1].
(tm_if t1' t2 t3)...
(* FILL IN HERE *) Admitted.
This is more interesting than the strong progress theorem that we saw in the Smallstep chapter, where all normal forms were values. Here, a term can be stuck, but only if it is ill typed.

#### Exercise: 1 star (step_review)

Quick review. Answer true or false. (As usual, no need to hand these in.)
• In this language, every well-typed normal form is a value.
• Every value is a normal form.
• The single-step evaluation relation is a partial function.
• The single-step evaluation relation is a total function.

## Type Preservation

The second critical property of typing is that, when a well-typed term takes a step, the result is also a well-typed term.
This theorem is often called the subject reduction property, because it tells us what happens when the "subject" of the typing relation is reduced. This terminology comes from thinking of typing statements as sentences, where the term is the subject and the type is the predicate.

#### Exercise: 3 stars, recommended (finish_preservation_informal)

Complete the following proof:
Theorem: If t : T and t t', then t' : T.
Proof: By induction on a derivation of t : T.
• If the last rule in the derivation is T_If, then t = if t1 then t2 else t3, with t1 : Bool, t2 : T and t3 : T.
Inspecting the rules for the small-step reduction relation and remembering that t has the form if ..., we see that the only ones that could have been used to prove t t' are ST_IfTrue, ST_IfFalse, or ST_If.
• If the last rule was ST_IfTrue, then t' = t2. But we know that t2 : T, so we are done.
• If the last rule was ST_IfFalse, then t' = t3. But we know that t3 : T, so we are done.
• If the last rule was ST_If, then t' = if t1' then t2 else t3, where t1 t1'. We know t1 : Bool so, by the IH, t1' : Bool. The T_If rule then gives us if t1' then t2 else t3 : T, as required.
(* FILL IN HERE *)

#### Exercise: 2 stars (finish_preservation)

Theorem preservation : t t' T,
has_type t T
t t'
has_type t' T.
Proof with auto.
intros t t' T HT HE.
generalize dependent t'.
has_type_cases (induction HT) Case;
(* every case needs to introduce a couple of things *)
intros t' HE;
(* and we can deal with several impossible
cases all at once *)

try (solve by inversion).
Case "T_If". inversion HE; subst.
SCase "ST_IFTrue". assumption.
SCase "ST_IfFalse". assumption.
SCase "ST_If". apply T_If; try assumption.
apply IHHT1; assumption.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars (preservation_alternate_proof)

Now prove the same property again by induction on the evaluation derivation instead of on the typing derivation. Begin by carefully reading and thinking about the first few lines of the above proof to make sure you understand what each one is doing. The set-up for this proof is similar, but not exactly the same.

Theorem preservation' : t t' T,
has_type t T
t t'
has_type t' T.
Proof with eauto.
(* FILL IN HERE *) Admitted.

## Type Soundness

Putting progress and preservation together, we can see that a well-typed term can never reach a stuck state.

Definition stepmany := (refl_step_closure step).
Notation "t1 '⇒*' t2" := (stepmany t1 t2) (at level 40).

Corollary soundness : t t' T,
has_type t T
t ⇒* t'
~(stuck t').
Proof.
intros t t' T HT P. induction P; intros [R S].
destruct (progress x T HT); auto.
apply IHP. apply (preservation x y T HT H).
unfold stuck. split; auto. Qed.

Indeed, in the present — extremely simple — language, every well-typed term can be reduced to a normal form: this is the normalization property. (The proof is straightforward, though somewhat tedious.) In richer languages, this property often fails, though there are some interesting languages (such as Coq's Fixpoint language, and the simply typed lambda-calculus, which we'll be looking at next) where all well-typed terms can be reduced to normal forms.

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

Having seen the subject reduction property, it is reasonable to wonder whether the opposity property — subject expansion — also holds. 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 *)

#### Exercise: 2 stars (variation1)

Suppose we add the following two new rules to the evaluation relation:
| ST_PredTrue :
(tm_pred tm_true (tm_pred tm_false)
| ST_PredFalse :
(tm_pred tm_false (tm_pred tm_true)
Which of the following properties remain true in the presence of these rules? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample.
• Determinacy of step
• Normalization of step for well-typed terms
• Progress
• Preservation

#### Exercise: 2 stars (variation2)

Suppose, instead, that we add this new rule to the typing relation:
| T_IfFunny :  t2 t3,
has_type t2 ty_Nat
has_type (tm_if tm_true t2 t3ty_Nat
Which of the following properties remain true in the presence of this rule? (Answer in the same style as above.)
• Determinacy of step
• Normalization of step for well-typed terms
• Progress
• Preservation

#### Exercise: 2 stars (variation3)

Suppose, instead, that we add this new rule to the typing relation:
| T_SuccBool :  t,
has_type t ty_Bool
has_type (tm_succ tty_Bool
Which of the following properties remain true in the presence of this rule? (Answer in the same style as above.)
• Determinacy of step
• Normalization of step for well-typed terms
• Progress
• Preservation

#### Exercise: 2 stars (variation4)

Suppose, instead, that we add this new rule to the step relation:
| ST_Funny1 :  t2 t3,
(tm_if tm_true t2 t3 t3
Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example.

#### Exercise: 2 stars (variation5)

| ST_Funny2 :  t1 t2 t2' t3,
t2  t2'
(tm_if t1 t2 t3 (tm_if t1 t2' t3)
Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example.

#### Exercise: 2 stars (variation6)

| ST_Funny3 :
(tm_pred tm_false (tm_pred (tm_pred tm_false))
Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example.

#### Exercise: 2 stars (variation7)

| T_Funny4 :
has_type tm_zero ty_Bool
Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example.

#### Exercise: 2 stars (variation8)

| T_Funny5 :
has_type (tm_pred tm_zeroty_Bool
Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example.

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

Make up some exercises of your own along the same lines as the ones above. Try to find ways of selectively breaking properties — i.e., ways of changing the definitions that break just one of the properties and leave the others alone.

#### Exercise: 1 star (remove_predzero)

The evaluation rule E_PredZero is a bit counter-intuitive: we might feel that it makes more sense for the predecessor of zero to be undefined, rather than being defined to be zero. Can we achieve this simply by removing the rule from the definition of step?
(* FILL IN HERE *)

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

Suppose our evaluation relation is defined in the big-step style. What are the appropriate analogs of the progress and preservation properties?
(* FILL IN HERE *)

# 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 above when formalizing of 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):
```<
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 :=
| ty_Bool : ty
| ty_arrow : ty ty ty.

### Terms

Inductive tm : Type :=
| tm_var : id tm
| tm_app : tm tm tm
| tm_abs : id ty tm tm
| tm_true : tm
| tm_false : tm
| tm_if : tm tm tm tm.

Something to note here is that an abstraction \x:T.t (formally, tm_abs x T t) is always annotated with the type (T) of its parameter. This is in contrast to Coq (and other functional languages like ML, Haskell, etc.), which use type inference to fill in missing annotations.

Tactic Notation "tm_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "tm_var" | Case_aux c "tm_app"
| Case_aux c "tm_abs" | Case_aux c "tm_true"
| Case_aux c "tm_false" | Case_aux c "tm_if" ].

Some examples...

Notation a := (Id 0).
Notation b := (Id 1).
Notation c := (Id 2).

idB = \a:Bool. a

Notation idB :=
(tm_abs a ty_Bool (tm_var a)).

idBB = \a:BoolBool. a

Notation idBB :=
(tm_abs a (ty_arrow ty_Bool ty_Bool) (tm_var a)).

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

Notation idBBBB :=
(tm_abs a (ty_arrow (ty_arrow ty_Bool ty_Bool)
(ty_arrow ty_Bool ty_Bool))
(tm_var a)).

k = \a:Bool. \b:Bool. a

Notation k := (tm_abs a ty_Bool (tm_abs b ty_Bool (tm_var a))).

(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 \a:A.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 \a:A.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 makes the first choice — for example,
Eval simpl in (fun a:bool => 3 + 4)
yields fun a:bool => 7. But most real 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 (tm_abs x T t)
| t_true :
value tm_true
| t_false :
value tm_false.

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 [s/x]t and pronounced "substitute s for x in t."
Here are some examples:
• [true / a] (if a then a else false) yields if true then true else false
• [true / a] a yields true
• [true / a] (if a then a else b) yields if true then true else b
• [true / a] b yields b
• [true / a] false yields false (vacuous substitution)
• [true / a] (\y:Bool. if y then a else false) yields \y:Bool. if y then true else false
• [true / a] (\y:Bool. a) yields \y:Bool. true
• [true / a] (\y:Bool. y) yields \y:Bool. y
• [true / a] (\a:Bool. a) yields \a:Bool. a
The last example is very important: substituting true for a in \a:Bool. a does not yield \a:Bool. true! The reason for this is that the a in the body of \a:Bool. a is bound by the abstraction: it is a new, local name that just happens to be spelled the same as some global name a.
Here is the definition, informally...
[s/x]x = s
[s/x]y = y                                     if x <> y
[s/x](\x:T11.t12)   = \x:T11. t12
[s/x](\y:T11.t12)   = \y:T11. [s/x]t12         if x <> y
[s/x](t1 t2)        = ([s/x]t1) ([s/x]t2)
[s/x]true           = true
[s/x]false          = false
[s/x](if t1 then t2 else t3) =
if [s/x]t1 then [s/x]t2 else [s/x]t3
... and formally:

Fixpoint subst (s:tm) (x:id) (t:tm) : tm :=
match t with
| tm_var x' => if beq_id x x' then s else t
| tm_abs x' T t1 => tm_abs x' T (if beq_id x x' then t1 else (subst s x t1))
| tm_app t1 t2 => tm_app (subst s x t1) (subst s x t2)
| tm_true => tm_true
| tm_false => tm_false
| tm_if t1 t2 t3 => tm_if (subst s x t1) (subst s x t2) (subst s x t3)
end.

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 in defining the step relation on closed terms here, we can avoid this extra complexity.

### Reduction

The small-step reduction relation for STLC 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
(\a:T.t12v2  [v2/a]t12
Informally:
 (ST_AppAbs) (\a:T.t12) v2 ⇒ [v2/a]t12
 t1 ⇒ t1' (ST_App1) t1 t2 ⇒ t1' t2
 t2 ⇒ t2' (ST_App2) v1 t2 ⇒ v1 t2'
(plus the usual rules for booleans).
Formally:

Reserved Notation "t1 '' t2" (at level 40).

Inductive step : tm tm Prop :=
| ST_AppAbs : x T t12 v2,
value v2
(tm_app (tm_abs x T t12) v2) (subst v2 x t12)
| ST_App1 : t1 t1' t2,
t1 t1'
tm_app t1 t2 tm_app t1' t2
| ST_App2 : v1 t2 t2',
value v1
t2 t2'
tm_app v1 t2 tm_app v1 t2'
| ST_IfTrue : t1 t2,
(tm_if tm_true t1 t2) t1
| ST_IfFalse : t1 t2,
(tm_if tm_false t1 t2) t2
| ST_If : t1 t1' t2 t3,
t1 t1'
(tm_if t1 t2 t3) (tm_if 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" ].

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

Hint Constructors step.

### Examples

Lemma step_example1 :
(tm_app idBB idB) ⇒* idB.
Proof.
eapply rsc_step.
apply ST_AppAbs.
apply v_abs.
simpl.
apply rsc_refl. Qed.

(* A more automatic proof *)
Lemma step_example1' :
(tm_app idBB idB) ⇒* idB.
Proof. normalize. Qed.

Lemma step_example2 :
(tm_app idBB (tm_app idBB idB)) ⇒* idB.
Proof.
eapply rsc_step.
apply ST_App2. auto.
apply ST_AppAbs. auto.
eapply rsc_step.
apply ST_AppAbs. simpl. auto.
simpl. apply rsc_refl. Qed.

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

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

#### Exercise: 2 stars (step_example3)

Try to do this one both with and without normalize.

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

## 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.

Definition context := partial_map ty.

Module Context.

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

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

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 Context.

### Typing Relation

Informally:
 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
The notation Γ , x:T means "extend the partial function Γ to also map x to T."

Inductive has_type : context tm ty Prop :=
| T_Var : Γ x T,
Γ x = Some T
has_type Γ (tm_var x) T
| T_Abs : Γ x T11 T12 t12,
has_type (extend Γ x T11) t12 T12
has_type Γ (tm_abs x T11 t12) (ty_arrow T11 T12)
| T_App : T11 T12 Γ t1 t2,
has_type Γ t1 (ty_arrow T11 T12)
has_type Γ t2 T11
has_type Γ (tm_app t1 t2) T12
| T_True : Γ,
has_type Γ tm_true ty_Bool
| T_False : Γ,
has_type Γ tm_false ty_Bool
| T_If : t1 t2 t3 T Γ,
has_type Γ t1 ty_Bool
has_type Γ t2 T
has_type Γ t3 T
has_type Γ (tm_if 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 (tm_abs a ty_Bool (tm_var a)) (ty_arrow ty_Bool ty_Bool).
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 (tm_abs a ty_Bool (tm_var a)) (ty_arrow ty_Bool ty_Bool).
Proof. auto. Qed.

Hint Unfold beq_id beq_nat extend.

Written informally, the next one is:
empty  \a:A. \b:AA. b (b a))
: A  (AA A.

Example typing_example_2 :
has_type empty
(tm_abs a ty_Bool
(tm_abs b (ty_arrow ty_Bool ty_Bool)
(tm_app (tm_var b) (tm_app (tm_var b) (tm_var a)))))
(ty_arrow ty_Bool (ty_arrow (ty_arrow ty_Bool ty_Bool) ty_Bool)).
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

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

Example typing_example_2_full :
has_type empty
(tm_abs a ty_Bool
(tm_abs b (ty_arrow ty_Bool ty_Bool)
(tm_app (tm_var b) (tm_app (tm_var b) (tm_var a)))))
(ty_arrow ty_Bool (ty_arrow (ty_arrow ty_Bool ty_Bool) ty_Bool)).
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 2 stars (typing_example_3)

Formally prove the following typing derivation holds:
empty  (\a:BoolB. \b:BoolBool. \c:Bool.
b (a c))
: T.

Example typing_example_3 :
T,
has_type empty
(tm_abs a (ty_arrow ty_Bool ty_Bool)
(tm_abs b (ty_arrow ty_Bool ty_Bool)
(tm_abs c ty_Bool
(tm_app (tm_var b) (tm_app (tm_var a) (tm_var c))))))
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 \a:Bool. \b:Bool, a b — i.e.,
~  T,
empty  (\a:Bool. \b:Boola b) : T.
Example typing_nonexample_1 :
~ T,
has_type empty
(tm_abs a ty_Bool
(tm_abs b ty_Bool
(tm_app (tm_var a) (tm_var b))))
T.
Proof.
intros C. destruct C.
(* 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  (\a:S. a a) : T).

Example typing_nonexample_3 :
~ ( S, T,
has_type empty
(tm_abs a S
(tm_app (tm_var a) (tm_var a)))
T).
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 1 star (typing_statements)

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

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

Which of the following propositions are provable? For the ones that are, give witnesses for the existentially bound variables.
• T, empty (\b:BBB. \a:B, b a) : T
• T, empty (\a:AB, \b:B-->C, \c:A, b (a c)):T
• S, U, T, a:S, b:U \c:A. a (b c) : T
• S, T, a:S \b:A. a (a b) : T
• S, U, T, a:S a (\c:U. c a) : T

## Properties

### 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 (tm_var x)
| afi_app1 : x t1 t2,
appears_free_in x t1 appears_free_in x (tm_app t1 t2)
| afi_app2 : x t1 t2,
appears_free_in x t2 appears_free_in x (tm_app t1 t2)
| afi_abs : x y T11 t12,
y <> x
appears_free_in x t12
appears_free_in x (tm_abs y T11 t12)
| afi_if1 : x t1 t2 t3,
appears_free_in x t1
appears_free_in x (tm_if t1 t2 t3)
| afi_if2 : x t1 t2 t3,
appears_free_in x t2
appears_free_in x (tm_if t1 t2 t3)
| afi_if3 : x t1 t2 t3,
appears_free_in x t3
appears_free_in x (tm_if 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. 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 Gamma'. When is it safe to do this? Intuitively, it must at least be the case that Gamma' 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 Gamma' t S,
has_type Γ t S
( x, appears_free_in x t Γ x = Gamma' x)
has_type Gamma' t S.

Proof: By induction on a derivation of Γ t : T.
• If the last rule in the derivation was T_Var, then t = x and Γ x = T. By assumption, Gamma' x = T as well, and hence Gamma' 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 Gamma'', if Γ, y:T11 and Gamma'' assign the same types to all the free variables in t12, then t12 has type T12 under Gamma''. Let Gamma' be a context which agrees with Γ on the free variables in t; we must show Gamma' \y:T11. t12 : T11 T12.
By T_Abs, it suffices to show that Gamma', y:T11 t12 : T12. By the IH (setting Gamma'' = Gamma', y:T11), it suffices to show that Γ, y:T11 and Gamma', 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 Gamma', 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 Gamma' agree on all such variables, and hence so do Γ, y:T11 and Gamma', 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 Gamma', if Gamma' agrees with Γ on the free variables in t1, then t1 has type T2 T under Gamma'; there is a similar IH for t2. We must show that t1 t2 also has type T under Gamma', given the assumption that Gamma' 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 Gamma' 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 Gamma'.
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 x0 Hafi.
(* the only tricky step... the Gamma' we use to
instantiate is extend Γ x T11 *)

unfold extend. remember (beq_id x x0) 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 [v/x]t : T.

Lemma substitution_preserves_typing : Γ x U v t T,
has_type (extend Γ x U) t T
has_type empty v U
has_type Γ (subst v x t) T.

One technical subtlety in the statement of the lemma is that we assign v the type U in the empty context — in other words, we assume v is closed. This assumption considerably simplifies the T_Abs case of the proof (compared to assuming Γ v : U, which would be the other reasonable assumption at this point) because the context invariance lemma then tells us that v has type U in any context at all — we don't have to worry about free variables in v 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 v : U, then Γ [v/x]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 [v/x]x = v has type T under Γ, given the assumption that v 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 Gamma' and T', that if Gamma',x:U t12 : T' and v : U, then Gamma' [v/x]t12 : T'. In particular, if Γ,y:T11,x:U t12 : T12 and v : U, then Γ,y:T11 [v/x]t12 : T12. There are again two cases to consider, depending on whether x and y are the same variable name.
First, suppose x = y. Then, by the definition of substitution, [v/x]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 [v/x]t12 : T12. By T_Abs, Γ \y:T11. [v/x]t12 : T11T12, and by the definition of substitution (noting that x <> y), Γ \y:T11. [v/x]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 v t T Ht Hv.
generalize dependent Γ. generalize dependent T.
tm_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 "tm_var".
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 "tm_abs".
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 [v/x] 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 subst t2 x 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 auto 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 earlier in this file 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 *)

### Progress

Finally, the progress theorem tells us that closed, well-typed terms are never stuck: either a well-typed term is a value, or else 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. (subst t2 x t)...
solve by inversion. solve by inversion.
SSCase "t2 steps".
destruct H0 as [t2' Hstp]. (tm_app t1 t2')...

SCase "t1 steps".
destruct H as [t1' Hstp]. (tm_app 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".
destruct H as [t1' Hstp]. (tm_if 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 types.

Theorem progress' : t T,
has_type empty t T
value t t', t t'.
Proof.
intros t.
tm_cases (induction t) Case; intros T Ht; auto.
(* 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, optional (stlc_variation1)

Suppose we add the following new rule to the evaluation relation of the STLC:
| T_Strange :  x t,
has_type empty (tm_abs x Bool tBool
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.
• Determinacy of step
• Progress
• Preservation

#### Exercise: 2 stars (stlc_variation2)

Suppose we remove the rule ST_App1 from the step relation. Which of the three properties in the previous exercise become false in the absence of this rule? For each that becomes false, give a counterexample.

End STLC.

# 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 :=
| ty_arrow : ty ty ty
| ty_Nat : ty.

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

Inductive tm : Type :=
| tm_var : id tm
| tm_app : tm tm tm
| tm_abs : id ty tm tm
| tm_nat : nat tm
| tm_succ : tm tm
| tm_pred : tm tm
| tm_mult : tm tm tm
| tm_if0 : tm tm tm tm.

Tactic Notation "tm_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "tm_var" | Case_aux c "tm_app"
| Case_aux c "tm_abs" | Case_aux c "tm_nat"
| Case_aux c "tm_succ" | Case_aux c "tm_pred"
| Case_aux c "tm_mult" | Case_aux c "tm_if0" ].

#### Exercise: 4 stars, recommended (STLCArith)

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.