# Stlc: The Simply Typed Lambda-Calculus

# More Automation

## The auto and eauto Tactics

- intros
- apply (with a local hypothesis, by default)
- reflexivity

Lemma auto_example_1 : forall P Q R S T U : Prop,

(P -> Q) ->

(P -> R) ->

(T -> R) ->

(S -> T -> U) ->

((P->Q) -> (P->S)) ->

T ->

P ->

U.

Proof.

auto. Qed.

By default, auto and eauto consider just the hypotheses in the
current context. But we can also give a list of lemmas and
constructors to consider in addition to these, by writing auto
using ....

Lemma auto_example_2 : forall P Q R : Prop,

Q ->

(Q -> R) ->

P \/ (Q /\ R).

Proof.

auto using conj, or_introl, or_intror. Qed.

There are some constructors and lemmas that are applied very often
in proofs. We can tell auto to
It is also sometimes necessary to add
Here are some Hints we will find useful. There is an example of
the effect of the and and or hints below.

*always*consider these, instead of mentioning them explicitly each time. This is accomplished 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 add *all*of the constructors from the inductive definition of c to the database used by auto.
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.
Hint Constructors and.

Hint Constructors or.

Hint Constructors refl_step_closure.

Hint Resolve beq_id_eq beq_id_false_not_eq.

Lemma auto_example_3 : forall P Q R : Prop,

Q ->

(Q -> R) ->

P \/ (Q /\ R).

Proof.

auto. Qed.

Hint Constructors or.

Hint Constructors refl_step_closure.

Hint Resolve beq_id_eq beq_id_false_not_eq.

Lemma auto_example_3 : forall P Q R : Prop,

Q ->

(Q -> R) ->

P \/ (Q /\ R).

Proof.

auto. Qed.

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 more
sparingly.
If you start a proof with Proof with (tactic) 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.
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.
Unfortunately, Coq does not provide a built-in tactic that does
this. However, it is not too hard to define one
ourselves. (Thanks to Aaron Bohannon for this nice hack.)

## The Proof with Tactic

## The solve by inversion Tactic

Tactic Notation "solve_by_inversion_step" tactic(t) :=

match goal with

| H : _ |- _ => solve [ inversion H; subst; t ]

end

|| fail "because the goal is not solvable by inversion.".

Tactic Notation "solve" "by" "inversion" "1" :=

solve_by_inversion_step idtac.

Tactic Notation "solve" "by" "inversion" "2" :=

solve_by_inversion_step (solve by inversion 1).

Tactic Notation "solve" "by" "inversion" "3" :=

solve_by_inversion_step (solve by inversion 2).

Tactic Notation "solve" "by" "inversion" :=

solve by inversion 1.

Again, the details of how the Tactic Notation definition works
are not important. All we need to know is that 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.
If t is a tactic, then try solve [t] is a tactic that
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
When experimenting when the languages defined in this file, we
will sometimes want to see what a particular concrete term steps
to, so we want to solve goals of the form t -->* t' where
t is a completely concrete term. These proofs are simple but
repetitive. Consider for example reducing an arithmetic expression
using the small-step relation astep defined in the previous
chapter:

## The try solve Tactic

- if t solves the goal, behaves just like t, or
- if t cannot completely solve the goal, does nothing.

*not*leave any of the subgoals in a mangled state.## The normalize Tactic

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 Ltac 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.

Ltac print_goal := match goal with |- ?x => idtac x end.

Ltac 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''' : exists e',

(APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state -->a* e'.

Proof.

eapply ex_intro. normalize.

Qed.

(APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state -->a* e'.

Proof.

eapply ex_intro. normalize.

Qed.

# Typed Arithmetic Expressions

## 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 : forall t, nvalue t -> nvalue (tm_succ t).

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

Hint Constructors bvalue nvalue.

Hint Unfold value.

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

Inductive step : tm -> tm -> Prop :=

| ST_IfTrue : forall t1 t2,

(tm_if tm_true t1 t2) --> t1

| ST_IfFalse : forall t1 t2,

(tm_if tm_false t1 t2) --> t2

| ST_If : forall t1 t1' t2 t3,

t1 --> t1' ->

(tm_if t1 t2 t3) --> (tm_if t1' t2 t3)

| ST_Succ : forall t1 t1',

t1 --> t1' ->

(tm_succ t1) --> (tm_succ t1')

| ST_PredZero :

(tm_pred tm_zero) --> tm_zero

| ST_PredSucc : forall t1,

nvalue t1 ->

(tm_pred (tm_succ t1)) --> t1

| ST_Pred : forall t1 t1',

t1 --> t1' ->

(tm_pred t1) --> (tm_pred t1')

| ST_IszeroZero :

(tm_iszero tm_zero) --> tm_true

| ST_IszeroSucc : forall t1,

nvalue t1 ->

(tm_iszero (tm_succ t1)) --> tm_false

| ST_Iszero : forall t1 t1',

t1 --> t1' ->

(tm_iszero t1) --> (tm_iszero t1')

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

Tactic Notation "step_cases" tactic(first) tactic(c) :=

first;

[ c "ST_IfTrue" | c "ST_IfFalse" | c "ST_If"

| c "ST_Succ" | c "ST_PredZero" | c "ST_PredSucc" | c "ST_Pred"

| c "ST_IszeroZero" | c "ST_IszeroSucc" | c "ST_Iszero" ].

Hint Constructors step.

Notice that the step relation doesn't care at all about whether
expressions make sense or not. For example, it is easy to see
that
The first interesting thing about the step relation is that the
progress theorem as we stated it before fails! That is, there are
terms that are normal forms (they can't take a step) but not
values -- i.e., we have not included them in our definition of
possible "results of evaluation." Such terms are said to be
stuck.

tm_succ (tm_if tm_true tm_true tm_true)

is nonsensical, but the step relation will happily reduce it one
step to
tm_succ tm_true.

## Normal Forms and Values

Notation step_normal_form := (normal_form step).

Definition stuck (t:tm) : Prop :=

step_normal_form t /\ ~ value t.

Hint Unfold stuck.

☐
Fortunately, things are not

*completely*messed up: values and normal forms are not the same in this language, but at least the former set is included in the latter (i.e., 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.
☐

#### Exercise: 3 stars, optional

Using value_is_nf, we can show that the step relation is also deterministic...
☐
Critical observation: although there are stuck terms in this
language, they are all "nonsensical", mixing booleans and numbers
in a way that we don't even

## Typing

*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.
The typing relation
In informal notation, the typing relation is often written |- t :
T, pronounced "t has type T."

(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 : Not |

|- t1 : Nat | (T_Pred) |

|- pred t1 : Not |

|- 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 : forall 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 : forall t1,

has_type t1 ty_nat ->

has_type (tm_succ t1) ty_nat

| T_Pred : forall t1,

has_type t1 ty_nat ->

has_type (tm_pred t1) ty_nat

| T_Iszero : forall t1,

has_type t1 ty_nat ->

has_type (tm_iszero t1) ty_bool.

Tactic Notation "has_type_cases" tactic(first) tactic(c) :=

first;

[ c "T_True" | c "T_False" | c "T_If" | c "T_Zero" | c "T_Succ"

| c "T_Pred" | c "T_Iszero" ].

Hint Constructors has_type.

### Examples

*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.

Example has_type_not :

~ has_type (tm_if tm_false tm_zero tm_true) ty_bool.

Proof.

intros Contra. solve by inversion 2. Qed.

Example succ_hastype_nat__hastype_nat : forall t,

has_type (tm_succ t) ty_nat ->

has_type t ty_nat.

Proof.

(* FILL IN HERE *) Admitted.

has_type (tm_succ t) ty_nat ->

has_type t ty_nat.

Proof.

(* FILL IN HERE *) Admitted.

☐
The typing relation enjoys two critical properties. The first is
that well-typed normal forms are values.

## Progress

#### Exercise: 2 stars (finish_progress)

Theorem progress : forall t T,

has_type t T ->

value t \/ exists t', t --> t'.

Proof with eauto.

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".

exists t2...

SSSCase "t1 is tm_false".

exists 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].

exists (tm_if t1' t2 t3)...

(* FILL IN HERE *) Admitted.

has_type t T ->

value t \/ exists t', t --> t'.

Proof with eauto.

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".

exists t2...

SSSCase "t1 is tm_false".

exists 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].

exists (tm_if t1' t2 t3)...

(* FILL IN HERE *) Admitted.

☐
This is a bit more interesting than the progress theorem that we
saw in Smallstep.v, where
☐
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 also sometimes called the

*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

*subject reduction*property, because it tells us what happens when the "subject" of the typing relation is reduced.#### Exercise: 2 stars (finish_preservation)

Theorem preservation : forall t t' T,

has_type t T ->

t --> t' ->

has_type t' T.

Proof with eauto.

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.

has_type t T ->

t --> t' ->

has_type t' T.

Proof with eauto.

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, optional (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' : forall t t' T,

has_type t T ->

t --> t' ->

has_type t' T.

Proof with eauto.

(* FILL IN HERE *) Admitted.

☐
Putting progress and preservation together, we can see that a
well-typed term can

## Type Soundness

*never*reach a stuck state.Definition stepmany := (refl_step_closure step).

Notation "t1 '-->*' t2" := (stepmany t1 t2) (at level 40).

Corollary soundness : forall 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. 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
(* FILL IN HERE *)

☐
☐
☐
☐
☐
☐
☐
☐
☐
(* FILL IN HERE *)

☐
(* FILL IN HERE *)

☐
The simply typed lambda-calculus (STLC) can be described as the
simplest possible functional programming language -- it contains
anonymous functions, application, variables, and nothing else.
Since it is essentially a small fragment of Coq's built-in
functional language, we could informally use the same
notation for programs. For example,
However, this "fun ... => ..." notation is a little heavy,
especially for writing on the blackboard. So when we discuss
terms informally, we'll use another common way of writing function
abstractions in this language: \x:A.x, where \ stands for the
greek letter "lambda".
To keep things simple, we'll start with the
The main new technical issues that we'll have to face in this
section are dealing with

*well-typed*terms can be reduced to normal forms.## Additional exercises

#### Exercise: 2 stars (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.☐

#### 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.
(tm_pred tm_true) --> (tm_pred tm_false)

| ST_PredFalse :

(tm_pred tm_false) --> (tm_pred tm_true)

- 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 : forall t2 t3,

has_type t2 ty_nat ->

has_type (tm_if tm_true t2 t3) ty_nat

Which of the following properties remain true in the presence of
this rule? (Answer in the same style as above.)
has_type t2 ty_nat ->

has_type (tm_if tm_true t2 t3) ty_nat

- 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 : forall t,

has_type t ty_bool ->

has_type (tm_succ t) ty_bool

Which of the following properties remain true in the presence of
this rule? (Answer in the same style as above.)
has_type t ty_bool ->

has_type (tm_succ t) ty_bool

- 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 : forall 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.
(tm_if tm_true t2 t3) --> t3

#### Exercise: 2 stars (variation5)

Suppose instead that we add this rule:
| ST_Funny2 : forall 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.
t2 --> t2' ->

(tm_if t1 t2 t3) --> (tm_if t1 t2' t3)

#### Exercise: 2 stars (variation6)

Suppose instead that we add this rule:
| 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.
(tm_pred tm_false) --> (tm_pred (tm_pred tm_false))

#### Exercise: 2 stars (variation7)

Suppose instead that we add this rule:
| 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.
has_type tm_zero ty_bool

#### Exercise: 2 stars (variation8)

Suppose instead that we add this rule:
| T_Funny5 :

has_type (tm_pred 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.
has_type (tm_pred tm_zero) ty_bool

#### 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?☐

#### Exercise: 3 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?☐

# The Simply Typed Lambda-Calculus

fun x:A => x

is the identity function at the type A, and
fun x:A => fun y:B => x

is a function that takes an argument of type A, throws it
away, and returns the constant-x function (of type B->A).
*pure*simply typed lambda-calculus, where base types like A, B, etc. are completely uninterpreted -- i.e., there are no constants belonging to base types and no primitive operations over base types. At the end of the chapter, we'll see how to extend the pure system to something closer to a real programming language by adding base types like numbers.*variable binding*and*substitution*.- We've already seen how to formalize a language with
variables (Imp). There, however, the variables were all global.
In STLC, we need to handle
*bound*variables because of function parameters. - Moreover, instead of just looking up global variables in the store, we'll need to reduce function applications by substituting arguments for parameters in function bodies.

## Syntax and Operational Semantics

### Types

Some base types for use in examples...

### Terms

Inductive tm : Type :=

| tm_var : id -> tm

| tm_app : tm -> tm -> tm

| tm_abs : id -> ty -> tm -> tm.

One important thing 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.), where we can
write either

fun x : A => t

or
fun x => t

and trust Coq to fill in an appropriate annotation in the latter
case.
Tactic Notation "tm_cases" tactic(first) tactic(c) :=

first;

[ c "tm_var" | c "tm_app" | c "tm_abs" ].

Some variables for examples...

Some identity functions on various types...
id_A = \a:A. a

id_AarrowA = \a:A->A. a

id_AarrowA_arrow_AarrowA = \a:(A->A)->(A->A). a

Notation id_AarrowA_arrow_AarrowA :=

(tm_abs a (ty_arrow (ty_arrow A A) (ty_arrow A A)) (tm_var a)).

A function that takes an A and returns a constant function
of type B->A:
k = \a:A. \b:B. a

(We make these Notations rather than Definitions to make things
easier for auto.)
When we come to defining the values of the STLC, we have to
think a little.
First, an application is clearly
For abstractions, on the other hand, we have a choice:
Coq makes the first choice -- for example,
Having decided this, 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).

### Values

*not*a value: It represents a function being invoked on some argument, and the term cannot be finished computing till this invocation has actually happened.- We can say that \a:A.t is a value only if t 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.t is always a value, no matter whether t is one or not -- in other words, we can say that reduction stops at abstractions.

Eval simpl in (fun a:bool => plus 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 latter choice here.
Inductive value : tm -> Prop :=

| v_abs : forall x T t,

value (tm_abs x T t).

Hint Constructors value.

### Substitution

[(\b:A.b)/a]a = \b:A.b

[(\b:A.b)/a]c = c

[(\b:A.b)/a](\c:B. a) = \c:B. \b:A.b

[(\b:A.b)/a](\c:(A->A)->(A->A). c a)

= \c:(A->A)->(A->A). c (\b:A.b)

[(\b:A.b)/a](\a:B.a) = \a:B.a

[(\b:(B->B)->(B->B).b)/a](a (\a:B.a))

= (\b:(B->B)->(B->B).b) (\a:B.a)

Note that, in the last two examples, we do [(\b:A.b)/a]c = c

[(\b:A.b)/a](\c:B. a) = \c:B. \b:A.b

[(\b:A.b)/a](\c:(A->A)->(A->A). c a)

= \c:(A->A)->(A->A). c (\b:A.b)

[(\b:A.b)/a](\a:B.a) = \a:B.a

[(\b:(B->B)->(B->B).b)/a](a (\a:B.a))

= (\b:(B->B)->(B->B).b) (\a:B.a)

*not*substitute for the a in the body of an abstraction whose bound variable is named a.

Fixpoint subst (x:id) (s:tm) (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 x s t1))

| tm_app t1 t2 => tm_app (subst x s t1) (subst x s t2)

end.

Technical note: Substitution becomes trickier to define if we
consider the case where s, the term being substituted in, 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.
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
Informally:

Formally:

### Reduction

(\a:T.t12) v2 --> [v2/a]t12

is traditionally called "beta-reduction".
value v2 | (ST_AppAbs) |

(\a:T.t12) v2 --> [v2/a]t12 |

t1 --> t1' | (ST_App1) |

t1 t2 --> t1' t2 |

value v1 t2 --> t2' | (ST_App2) |

v1 t2 --> v1 t2' |

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

Inductive step : tm -> tm -> Prop :=

| ST_AppAbs : forall x T t12 v2,

value v2 ->

(tm_app (tm_abs x T t12) v2) --> (subst x v2 t12)

| ST_App1 : forall t1 t1' t2,

t1 --> t1' ->

tm_app t1 t2 --> tm_app t1' t2

| ST_App2 : forall v1 t2 t2',

value v1 ->

t2 --> t2' ->

tm_app v1 t2 --> tm_app v1 t2'

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

Tactic Notation "step_cases" tactic(first) tactic(c) :=

first;

[ c "ST_AppAbs" | c "ST_App1" | c "ST_App2" ].

Notation stepmany := (refl_step_closure step).

Notation "t1 '-->*' t2" := (stepmany t1 t2) (at level 40).

Hint Constructors step.

Lemma step_example1 :

(tm_app id_AarrowA id_A) -->* id_A.

Proof.

eapply rsc_step.

apply ST_AppAbs.

apply v_abs.

simpl.

apply rsc_refl. Qed.

Lemma step_example2 :

(tm_app id_AarrowA (tm_app id_AarrowA id_A)) -->* id_A.

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.

We can use the normalize tactic from above to simplify the
proof.

Lemma step_example2' :

(tm_app id_AarrowA (tm_app id_AarrowA id_A)) -->* id_A.

Proof.

normalize.

Qed.

Lemma step_example3 :

(tm_app (tm_app id_AarrowA_arrow_AarrowA id_AarrowA) id_A)

-->* id_A.

Proof.

(* FILL IN HERE *) Admitted.

(tm_app (tm_app id_AarrowA_arrow_AarrowA id_AarrowA) id_A)

-->* id_A.

Proof.

(* FILL IN HERE *) Admitted.

☐
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 Gamma |- t : T, where Gamma is a "typing context"
a mapping from variables to their types.

## Typing

### Contexts

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

Definition context := partial_map ty.

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

Definition extend {A:Type} (Gamma : partial_map A) (x:id) (T : A) :=

fun x' => if beq_id x x' then Some T else Gamma x'.

Lemma extend_eq : forall 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 : forall 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.

### Typing Relation

Gamma x = T | (T_Var) |

Gamma |- x : T |

Gamma, x:T11 |- t12 : T12 | (T_Abs) |

Gamma |- \x:T11.t12 : T11->T12 |

Gamma |- t1 : T11->T12 | |

Gamma |- t2 : T11 | (T_App) |

Gamma |- t1 t2 : T12 |

Inductive has_type : context -> tm -> ty -> Prop :=

| T_Var : forall Gamma x T,

Gamma x = Some T ->

has_type Gamma (tm_var x) T

| T_Abs : forall Gamma x T11 T12 t12,

has_type (extend Gamma x T11) t12 T12 ->

has_type Gamma (tm_abs x T11 t12) (ty_arrow T11 T12)

| T_App : forall T11 T12 Gamma t1 t2,

has_type Gamma t1 (ty_arrow T11 T12) ->

has_type Gamma t2 T11 ->

has_type Gamma (tm_app t1 t2) T12.

Tactic Notation "typing_cases" tactic(first) tactic(c) :=

first;

[ c "T_Var" | c "T_Abs" | c "T_App" ].

Hint Constructors has_type.

Example typing_example_1 :

has_type empty (tm_abs a A (tm_var a)) (ty_arrow A A).

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 A (tm_var a)) (ty_arrow A A).

Proof. auto. Qed.

Hint Unfold beq_id beq_nat extend.

Written informally, the next one is:

empty |- \a:A. \b:A->A. b (b a))

: A -> (A->A) -> A.

: A -> (A->A) -> A.

Example typing_example_2 :

has_type empty

(tm_abs a A

(tm_abs b (ty_arrow A A)

(tm_app (tm_var b) (tm_app (tm_var b) (tm_var a)))))

(ty_arrow A (ty_arrow (ty_arrow A A) A)).

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.

Example typing_example_2_full :

has_type empty

(tm_abs a A

(tm_abs b (ty_arrow A A)

(tm_app (tm_var b) (tm_app (tm_var b) (tm_var a)))))

(ty_arrow A (ty_arrow (ty_arrow A A) A)).

Proof.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 2 stars (typing_example_3)

Formally prove the following typing derivation holds:
empty |- (\a:A->B. \b:B-->C. \c:A.

b (a c))

: T.

b (a c))

: T.

Example typing_example_3 :

exists T,

has_type empty

(tm_abs a (ty_arrow A B)

(tm_abs b (ty_arrow B C)

(tm_abs c A

(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:A. \b:B, a b -- i.e.,
~ exists T,

empty |- (\a:A. \b:B, a b) : T.

empty |- (\a:A. \b:B, a b) : T.

Example typing_nonexample_1 :

~ exists T,

has_type empty

(tm_abs a A

(tm_abs b B

(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.

~ exists T,

has_type empty

(tm_abs a A

(tm_abs b B

(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: 2 stars (typing_nonexample_2)

Now you prove this one doesn't hold:
~ exists T,

empty |- (\a:A->A, \b:B, a b) : T.

empty |- (\a:A->A, \b:B, a b) : T.

Example typing_nonexample_2 :

~ exists T,

has_type empty

(tm_abs a (ty_arrow A A)

(tm_abs b B

(tm_app (tm_var a) (tm_var b))))

T.

Proof.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 3 stars (typing_nonexample_3)

Another nonexample:
~ (exists S, exists T,

empty |- (\a:S. a a) : T).

empty |- (\a:S. a a) : T).

Example typing_nonexample_3 :

~ (exists S, exists T,

has_type empty

(tm_abs a S

(tm_app (tm_var a) (tm_var a)))

T).

Proof.

(* FILL IN HERE *) Admitted.

☐
Which of the following propositions are provable?
☐
Which of the following propositions are provable? For the
ones that are, give witnesses for the existentially bound
variables.
☐

#### Exercise: 1 star (typing_statements)

- b:B |- \a:A.a : A->A
- exists T, empty |- (\b:B->B. \a:B. b a) : T
- exists T, empty |- (\b:B->B. \a:B. a b) : T
- exists S, a:S |- (\b:B->B. b) a : S
- exists S, exists T, a:S |- (a a a) : T

#### Exercise: 1 star (more_typing_statements)

- exists T, empty |- (\b:B->B->B. \a:B, b a) : T
- exists T, empty |- (\a:A->B, \b:B-->C, \c:A, b (a c)):T
- exists S, exists U, exists T, a:S, b:U |- (\c:A. a b c) : T
- exists S, exists T, a:S |- \b:A. a (a b) : T
- exists S, exists U, exists T, a:S |- a (\c:U. c a) : T

## Properties

### Free Variables

Inductive appears_free_in : id -> tm -> Prop :=

| afi_var : forall x,

appears_free_in x (tm_var x)

| afi_app1 : forall x t1 t2,

appears_free_in x t1 -> appears_free_in x (tm_app t1 t2)

| afi_app2 : forall x t1 t2,

appears_free_in x t2 -> appears_free_in x (tm_app t1 t2)

| afi_abs : forall x y T11 t12,

y <> x ->

appears_free_in x t12 ->

appears_free_in x (tm_abs y T11 t12).

Tactic Notation "afi_cases" tactic(first) tactic(c) :=

first;

[ c "afi_var" | c "afi_app1" | c "afi_app2" | c "afi_abs" ].

Hint Constructors appears_free_in.

Definition closed (t:tm) :=

forall x, ~ appears_free_in x t.

Lemma free_in_context : forall x t T Gamma,

appears_free_in x t ->

has_type Gamma t T ->

exists T', Gamma x = Some T'.

Proof.

intros. generalize dependent Gamma. generalize dependent T.

(afi_cases (induction H) Case); intros.

Case "afi_var".

inversion H0; subst. exists T. assumption.

Case "afi_app1".

inversion H0; subst. eapply IHappears_free_in. apply H4.

Case "afi_app2".

inversion H0; subst. eapply IHappears_free_in. apply H6.

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.

Corollary typable_empty__closed : forall t T,

has_type empty t T ->

closed t.

Proof.

(* FILL IN HERE *) Admitted.

has_type empty t T ->

closed t.

Proof.

(* FILL IN HERE *) Admitted.

☐

Lemma context_invariance : forall Gamma Gamma' t S,

has_type Gamma t S ->

(forall x, appears_free_in x t -> Gamma x = Gamma' x) ->

has_type Gamma' t S.

Proof with eauto.

intros.

generalize dependent Gamma'.

(typing_cases (induction H) Case); intros.

Case "T_Var".

apply T_Var. rewrite <- H0...

Case "T_Abs".

apply T_Abs.

apply IHhas_type. intros x0 Hafi.

(* tricky step... the Gamma' we use to instantiate is

extend ty Gamma x T1 *)

unfold extend. remember (beq_id x x0) as e. destruct e...

Case "T_App".

apply T_App with T11... Qed.

Lemma substitution_preserves_typing : forall Gamma x U v t S,

has_type (extend Gamma x U) t S ->

has_type empty v U ->

has_type Gamma (subst x v t) S.

Proof with eauto.

intros Gamma x U v t S Ht Hv.

generalize dependent Gamma. generalize dependent S.

(tm_cases (induction t) Case); intros S Gamma 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 _ _ S 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.

Theorem preservation : forall t t' T,

has_type empty t T ->

t --> t' ->

has_type empty t' T.

Proof with eauto.

remember (@empty ty) as Gamma.

intros t t' T HT. generalize dependent t'.

(typing_cases (induction HT) Case); intros t' HE; subst Gamma.

Case "T_Var".

inversion HE.

Case "T_Abs".

inversion HE.

Case "T_App".

(step_cases (inversion HE) SCase); subst...

(* The ST_App1 and ST_App2 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 (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.☐

### Progress

Theorem progress : forall t T,

has_type empty t T ->

value t \/ exists t', t --> t'.

Proof with eauto.

intros t T Ht.

remember (@empty ty) as Gamma.

(typing_cases (induction Ht) Case); subst Gamma...

Case "T_Var".

inversion H.

Case "T_App".

right. destruct IHHt1...

SCase "t1 is a value".

destruct IHHt2...

SSCase "t2 is a value".

(* Since t2 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. exists (subst x t2 t)...

SSCase "t2 steps".

destruct H0 as [t2' Hstp]. exists (tm_app t1 t2')...

SCase "t1 steps".

destruct H as [t1' Hstp]. exists (tm_app t1' t2)...

Qed.

#### Exercise: 3 stars, optional

Show that progress can also be proved by induction on terms instead of types.Theorem progress' : forall t T,

has_type empty t T ->

value t \/ exists t', t --> t'.

Proof.

intros t.

(tm_cases (induction t) Case); intros T Ht.

(* FILL IN HERE *) Admitted.

☐

### Uniqueness of types

#### Exercise: 3 stars

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 *)

☐
☐
☐

## Additional exercises

#### 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 the following new rule to the evaluation relation of the STLC:
| T_Strange : forall x t,

has_type empty (tm_abs x A t) B

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.
has_type empty (tm_abs x A t) B

- 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.# Exercise: STLC with Arithmetic

## Syntax and operational semantics

To terms, we add natural number constants, along with
successor, predecessor, 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) tactic(c) :=

first;

[ c "tm_var" | c "tm_app" | c "tm_abs" |

c "tm_nat" | c "tm_succ" | c "tm_pred" | c "tm_mult" | c "tm_if0" ].

#### Exercise: 4 stars (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 *)

☐

Version of 4/26/2010