# StlcThe Simply Typed Lambda-Calculus

(* $Date: 2011-06-09 16:11:42 -0400 (Thu, 09 Jun 2011) $ *)

Require Export Types.

# The Simply Typed Lambda-Calculus

*functional abstraction*, which shows up in pretty much every real-world programming language in some form (functions, procedures, methods, etc.).

*variable binding*and

*substitution*.

## Overview

*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.

- variables
- function abstractions
- application

< t ::= x variable | \x:T.t1 abstraction | t1 t2 application | true constant true | false constant false | if t1 then t2 else t3 conditionalThe \ 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.

- \x:Bool. x
- (\x:Bool. x) true
- \x:Bool. if x then false else true
- \x:Bool. true
- \x:Bool. \y:Bool. x
- (\x:Bool. \y:Bool. x) false true
- \f:Bool→Bool. f (f true)
*function*f (from booleans to booleans) as an argument, applies f to true, and applies f again to the result. - (\f:Bool→Bool. f (f true)) (\x:Bool. false)

*higher-order*functions: we can write down functions that take other functions as arguments and/or return other functions as results.

*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.

*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 -> T2For example:

- \x:Bool. false has type Bool→Bool
- \x:Bool. x has type Bool→Bool
- (\x:Bool. x) true has type Bool
- \x:Bool. \y:Bool. x has type Bool→Bool→Bool (i.e. Bool → (Bool→Bool))
- (\x:Bool. \y:Bool. x) false has type Bool→Bool
- (\x:Bool. \y:Bool. x) false true has type Bool
- \f:Bool→Bool. f (f true) has type (Bool→Bool) → Bool
- (\f:Bool→Bool. f (f true)) (\x:Bool. false) has type Bool

## Syntax

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...

idB = \a:Bool. a

idBB = \a:Bool→Bool. a

idBBBB = \a:(Bool→Bool)->(Bool→Bool). 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

(We write these as Notations rather than Definitions to make
things easier for auto.)
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
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:
Coq makes the first choice — for example,
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).

## Operational Semantics

*free variables*and*substitution*, which are used in the reduction rule for application expressions. And finally we give the small-step relation itself.### Values

- 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.

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.
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

(\x:Bool. if x then true else x) false

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."
- [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

*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.

[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:
[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

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.
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:

(plus the usual rules for booleans).
Formally:

### Reduction

(\a:T.t12) v2 ⇒ [v2/a]t12

is traditionally called "beta-reduction".
(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' |

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.

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.

☐
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.

## Typing

### Contexts

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

Γ x = T | (T_Var) |

Γ ⊢ x : T |

Γ , x:T11 ⊢ t12 : T12 | (T_Abs) |

Γ ⊢ \x:T11.t12 : T11->T12 |

Γ ⊢ t1 : T11->T12 | |

Γ ⊢ t2 : T11 | (T_App) |

Γ ⊢ t1 t2 : T12 |

(T_True) | |

Γ ⊢ true : Bool |

(T_False) | |

Γ ⊢ false : Bool |

Γ ⊢ t1 : Bool Γ ⊢ t2 : T Γ ⊢ t3 : T | (T_If) |

Γ ⊢ if t1 then t2 else t3 : T |

Inductive has_type : context → tm → ty → Prop :=

| T_Var : ∀ Γ x T,

Γ x = Some T →

has_type Γ (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.

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:A→A. b (b a))

: A → (A→A) → A.

: A → (A→A) → 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.

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:Bool→B. \b:Bool→Bool. \c:Bool.

b (a c))

: T.

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:Bool, a b) : T.

empty ⊢ (\a:Bool. \b:Bool, a 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.

~ ∃ 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.

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.

☐
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.
☐
A variable x

#### Exercise: 1 star (typing_statements)

- b:Bool ⊢ \a:Bool.a : Bool→Bool
- ∃ T, empty ⊢ (\b:Bool→Bool. \a:Bool. b a) : T
- ∃ T, empty ⊢ (\b:Bool→Bool. \a:Bool. a b) : T
- ∃ S, a:S ⊢ (\b:Bool→Bool. b) a : S
- ∃ S, ∃ T, a:S ⊢ (a a a) : T

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

- ∃ T, empty ⊢ (\b:B→B→B. \a:B, b a) : T
- ∃ T, empty ⊢ (\a:A→B, \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

*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:T→U. x y
- both x and y appear free in (\x:T→U. x y) x
- no variables appear free in \x:T→U. \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*.### Substitution

*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.

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.
- 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
Formally, the so-called

*substitution*preserves types.*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
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,

*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 = 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 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.
- 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.

*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.
We now have the tools we need to prove

### Preservation

*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.*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 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.
- 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.

- If t steps to t2 or t3, the result is immediate, since
t2 and t3 have the same type as t.

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.☐

### Progress

*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.

*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.

- 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.
- If t1 can take a step, then so can t1 t2 by ST_App1.

- 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.
- 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).

- 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.

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 *)

☐
☐
☐

## 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, 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 t) Bool

Which of the following properties of the STLC remain true in
the presence of this rule? For each one, write either
"remains true" or else "becomes false." If a property becomes
false, give a counterexample.
has_type empty (tm_abs x Bool t) Bool

- 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, 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 *)

☐