# NormNormalization of STLC

(* \$Date: 2011-04-13 14:34:47 -0400 (Wed, 13 Apr 2011) \$ *)
(* Chapter maintained by Andrew Tolmach *)

(* (Based on TAPL Ch. 12.) *)

Require Import Stlc.
Require Import Relations.

(This chapter is optional.)
In this chapter, we consider another fundamental theoretical property of the simply typed lambda-calculus: the fact that the evaluation of a well-typed program is guaranteed to halt in a finite number of steps—-i.e., every well-typed term is normalizable.
Unlike the type-safety properties we have considered so far, the normalization property does not extend to full-blown programming languages, because these languages nearly always extend the simply typed lambda-calculus with constructs, such as general recursion (as we discussed in the MoreStlc chapter) or recursive types, that can be used to write nonterminating programs. However, the issue of normalization reappears at the level of types when we consider the metatheory of polymorphic versions of the lambda calculus such as F_omega: in this system, the language of types effectively contains a copy of the simply typed lambda-calculus, and the termination of the typechecking algorithm will hinge on the fact that a ``normalization'' operation on type expressions is guaranteed to terminate.
Another reason for studying normalization proofs is that they are some of the most beautiful—-and mind-blowing—-mathematics to be found in the type theory literature, often (as here) involving the fundamental proof technique of logical relations.
The calculus we shall consider here is the simply typed lambda-calculus over a single base type bool and with pairs. We'll give full details of the development for the basic lambda-calculus terms treating bool as an uninterpreted base type, and leave the extension to the boolean operators and pairs to the reader. Even for the base calculus, normalization is not entirely trivial to prove, since each reduction of a term can duplicate redexes in subterms.

#### Exercise: 1 star

Where do we fail if we attempt to prove normalization by a straightforward induction on the size of a well-typed term?

(* FILL IN HERE *)

# Language

We begin by repeating the relevant language definition, which is similar to those in the MoreStlc chapter, and supporting results including type preservation and step determinism. (We won't need progress.) You may just wish to skip down to the Normalization section...

### Syntax and Operational Semantics

Inductive ty : Type :=
| ty_Bool : ty
| ty_arrow : ty ty ty
| ty_prod : ty ty ty
.

Tactic Notation "ty_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "ty_Bool" | Case_aux c "ty_arrow" | Case_aux c "ty_prod" ].

Inductive tm : Type :=
(* pure STLC *)
| tm_var : id tm
| tm_app : tm tm tm
| tm_abs : id ty tm tm
(* pairs *)
| tm_pair : tm tm tm
| tm_fst : tm tm
| tm_snd : tm tm
(* booleans *)
| tm_true : tm
| tm_false : tm
| tm_if : tm tm tm tm.
(* i.e., if t0 then t1 else t2 *)

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_pair" | Case_aux c "tm_fst" | Case_aux c "tm_snd"
| Case_aux c "tm_true" | Case_aux c "tm_false" | Case_aux c "tm_if" ].

### Substitution

Fixpoint subst (x:id) (s:tm) (t:tm) : tm :=
match t with
| tm_var y => if beq_id x y then s else t
| tm_abs y T t1 => tm_abs y T (if beq_id x y then t1 else (subst x s t1))
| tm_app t1 t2 => tm_app (subst x s t1) (subst x s t2)
| tm_pair t1 t2 => tm_pair (subst x s t1) (subst x s t2)
| tm_fst t1 => tm_fst (subst x s t1)
| tm_snd t1 => tm_snd (subst x s t1)
| tm_true => tm_true
| tm_false => tm_false
| tm_if t0 t1 t2 => tm_if (subst x s t0) (subst x s t1) (subst x s t2)
end.

### Reduction

Inductive value : tm Prop :=
| v_abs : x T11 t12,
value (tm_abs x T11 t12)
| v_pair : v1 v2,
value v1
value v2
value (tm_pair v1 v2)
| v_true : value tm_true
| v_false : value tm_false
.

Hint Constructors value.

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

Inductive step : tm tm Prop :=
| ST_AppAbs : x T11 t12 v2,
value v2
(tm_app (tm_abs x T11 t12) v2) (subst x v2 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')
(* pairs *)
| ST_Pair1 : t1 t1' t2,
t1 t1'
(tm_pair t1 t2) (tm_pair t1' t2)
| ST_Pair2 : v1 t2 t2',
value v1
t2 t2'
(tm_pair v1 t2) (tm_pair v1 t2')
| ST_Fst : t1 t1',
t1 t1'
(tm_fst t1) (tm_fst t1')
| ST_FstPair : v1 v2,
value v1
value v2
(tm_fst (tm_pair v1 v2)) v1
| ST_Snd : t1 t1',
t1 t1'
(tm_snd t1) (tm_snd t1')
| ST_SndPair : v1 v2,
value v1
value v2
(tm_snd (tm_pair v1 v2)) v2
(* booleans *)
| ST_IfTrue : t1 t2,
(tm_if tm_true t1 t2) t1
| ST_IfFalse : t1 t2,
(tm_if tm_false t1 t2) t2
| ST_If : t0 t0' t1 t2,
t0 t0'
(tm_if t0 t1 t2) (tm_if t0' t1 t2)

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_Pair1" | Case_aux c "ST_Pair2"
| Case_aux c "ST_Fst" | Case_aux c "ST_FstPair"
| Case_aux c "ST_Snd" | Case_aux c "ST_SndPair"
| 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.

Notation step_normal_form := (normal_form step).

Lemma value__normal : t, value t step_normal_form t.
Proof with eauto.
intros t H; induction H; intros [t' ST]; inversion ST...
Qed.

### Typing

Definition context := partial_map ty.

Inductive has_type : context tm ty Prop :=
(* Typing rules for proper terms *)
| 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 : T1 T2 Γ t1 t2,
has_type Γ t1 (ty_arrow T1 T2)
has_type Γ t2 T1
has_type Γ (tm_app t1 t2) T2
(* pairs *)
| T_Pair : Γ t1 t2 T1 T2,
has_type Γ t1 T1
has_type Γ t2 T2
has_type Γ (tm_pair t1 t2) (ty_prod T1 T2)
| T_Fst : Γ t T1 T2,
has_type Γ t (ty_prod T1 T2)
has_type Γ (tm_fst t) T1
| T_Snd : Γ t T1 T2,
has_type Γ t (ty_prod T1 T2)
has_type Γ (tm_snd t) T2
(* booleans *)
| T_True : Γ,
has_type Γ tm_true ty_Bool
| T_False : Γ,
has_type Γ tm_false ty_Bool
| T_If : Γ t0 t1 t2 T,
has_type Γ t0 ty_Bool
has_type Γ t1 T
has_type Γ t2 T
has_type Γ (tm_if t0 t1 t2) T
.

Hint Constructors has_type.

Tactic Notation "has_type_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "T_Var" | Case_aux c "T_Abs" | Case_aux c "T_App"
| Case_aux c "T_Pair" | Case_aux c "T_Fst" | Case_aux c "T_Snd"
| Case_aux c "T_True" | Case_aux c "T_False" | Case_aux c "T_If" ].

Hint Extern 2 (has_type _ (tm_app _ _) _) => eapply T_App; auto.
Hint Extern 2 (_ = _) => compute; reflexivity.

### Context Invariance

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)
(* pairs *)
| afi_pair1 : x t1 t2,
appears_free_in x t1
appears_free_in x (tm_pair t1 t2)
| afi_pair2 : x t1 t2,
appears_free_in x t2
appears_free_in x (tm_pair t1 t2)
| afi_fst : x t,
appears_free_in x t
appears_free_in x (tm_fst t)
| afi_snd : x t,
appears_free_in x t
appears_free_in x (tm_snd t)
(* booleans *)
| afi_if0 : x t0 t1 t2,
appears_free_in x t0
appears_free_in x (tm_if t0 t1 t2)
| afi_if1 : x t0 t1 t2,
appears_free_in x t1
appears_free_in x (tm_if t0 t1 t2)
| afi_if2 : x t0 t1 t2,
appears_free_in x t2
appears_free_in x (tm_if t0 t1 t2)
.

Hint Constructors appears_free_in.

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

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 with eauto.
intros. generalize dependent Gamma'.
has_type_cases (induction H) Case;
intros Gamma' Heqv...
Case "T_Var".
apply T_Var... rewrite Heqv...
Case "T_Abs".
apply T_Abs... apply IHhas_type. intros y Hafi.
unfold extend. remember (beq_id x y) as e.
destruct e...
Case "T_Pair".
apply T_Pair...
Case "T_If".
eapply T_If...
Qed.

Lemma free_in_context : x t T Γ,
appears_free_in x t
has_type Γ t T
T', Γ x = Some T'.
Proof with eauto.
intros x t T Γ Hafi Htyp.
has_type_cases (induction Htyp) Case; inversion Hafi; subst...
Case "T_Abs".
destruct IHHtyp as [T' Hctx]... T'.
unfold extend in Hctx.
apply not_eq_beq_id_false in H2. rewrite H2 in Hctx...
Qed.

Corollary typable_empty__closed : t T,
has_type empty t T
closed t.
Proof.
intros. unfold closed. intros x H1.
destruct (free_in_context _ _ _ _ H1 H) as [T' C].
inversion C. Qed.

### Preservation

Lemma substitution_preserves_typing : Γ x U v t S,
has_type (extend Γ x U) t S
has_type empty v U
has_type Γ (subst x v t) S.
Proof with eauto.
(* Theorem: If Gamma,x:U |- t : S and empty |- v : U, then
Gamma |- (subst x v t) S. *)

intros Γ x U v t S Htypt Htypv.
generalize dependent Γ. generalize dependent S.
(* Proof: By induction on the term t.  Most cases follow directly
from the IH, with the exception of tm_var and tm_abs.
The former aren't automatic because we must reason about how the
variables interact. *)

tm_cases (induction t) Case;
intros S Γ Htypt; simpl; inversion Htypt; subst...
Case "tm_var".
simpl. rename i into y.
(* If t = y, we know that
empty v : U and
Γ,x:U y : S
and, by inversion, extend Γ x U y = Some S.  We want to
show that Γ subst x v y : S.

There are two cases to consider: either x=y or x<>y. *)

remember (beq_id x y) as e. destruct e.
SCase "x=y".
(* If x = y, then we know that U = S, and that subst x v y = v.
So what we really must show is that if empty v : U then
Γ v : U.  We have already proven a more general version
of this theorem, called context invariance. *)

apply beq_id_eq in Heqe. subst.
unfold extend in H1. rewrite beq_id_refl in H1.
inversion H1; subst. clear H1.
eapply context_invariance...
intros x Hcontra.
destruct (free_in_context _ _ S empty Hcontra) as [T' HT']...
inversion HT'.
SCase "x<>y".
(* If x <> y, then Γ y = Some S and the substitution has no
effect.  We can show that Γ y : S by T_Var. *)

apply T_Var... unfold extend in H1. rewrite Heqe in H1...
Case "tm_abs".
rename i into y. rename t into T11.
(* If t = tm_abs y T11 t0, then we know that
Γ,x:U tm_abs y T11 t0 : T11T12
Γ,x:U,y:T11 t0 : T12
empty v : U
As our IH, we know that forall S Gamma,
Γ,x:U t0 : S Γ subst x v t0 S.

We can calculate that
subst x v t = tm_abs y T11 (if beq_id x y
then t0
else subst x v t0)
And we must show that Γ subst x v t : T11T12.  We know
we will do so using T_Abs, so it remains to be shown that:
Γ,y:T11 if beq_id x y then t0 else subst x v t0 : T12
We consider two cases: x = y and x <> y.
*)

apply T_Abs...
remember (beq_id x y) as e. destruct e.
SCase "x=y".
(* If x = y, then the substitution has no effect.  Context
invariance shows that Γ,y:U,y:T11 and Γ,y:T11 are
equivalent.  Since the former context shows that t0 : T12, so
does the latter. *)

eapply context_invariance...
apply beq_id_eq in Heqe. subst.
intros x Hafi. unfold extend.
destruct (beq_id y x)...
SCase "x<>y".
(* If x <> y, then the IH and context invariance allow us to show that
Γ,x:U,y:T11 t0 : T12       =>
Γ,y:T11,x:U t0 : T12       =>
Γ,y:T11 subst x v t0 : T12 *)

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 : t t' T,
has_type empty t T
t t'
has_type empty t' T.
Proof with eauto.
intros t t' T HT.
(* Theorem: If empty t : T and t t', then empty t' : T. *)
remember (@empty ty) as Γ. generalize dependent HeqGamma.
generalize dependent t'.
(* Proof: By induction on the given typing derivation.  Many cases are
contradictory (T_VarT_Abs).  We show just the interesting ones. *)

has_type_cases (induction HT) Case;
intros t' HeqGamma HE; subst; inversion HE; subst...
Case "T_App".
(* If the last rule used was T_App, then t = t1 t2, and three rules
could have been used to show t t'ST_App1ST_App2, and
ST_AppAbs. In the first two cases, the result follows directly from
the IH. *)

inversion HE; subst...
SCase "ST_AppAbs".
(* For the third case, suppose
t1 = tm_abs x T11 t12
and
t2 = v2.
We must show that empty subst x v2 t12 : T2
We know by assumption that
empty tm_abs x T11 t12 : T1T2
and by inversion
x:T1 t12 : T2
We have already proven that substitution_preserves_typing and
empty v2 : T1
by assumption, so we are done. *)

apply substitution_preserves_typing with T1...
inversion HT1...
Case "T_Fst".
inversion HT...
Case "T_Snd".
inversion HT...
Qed.

### Determinism

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

# Normalization

Now for the actual normalization proof.
Our goal is to prove that every well-typed term evaluates to a normal form. In fact, it turns out to be convenient to prove something slightly stronger, namely that every well-typed term evaluates to a value. This follows from the weaker property anyway via the Progress lemma (why?) but otherwise we don't need Progress, and we didn't bother re-proving it above.
Here's the key definition:

Definition halts (t:tm) : Prop := t', t ⇒* t' value t'.

A trivial fact:

Lemma value_halts : v, value v halts v.
Proof.
intros v H. unfold halts.
v. split.
apply rsc_refl.
assumption.
Qed.

The key issue in the normalization proof (as in many proofs by induction) is finding a strong enough induction hypothesis. To this end, we begin by defining, for each type T, a set R_T of closed terms of type T. We will specify these sets using a relation R and write R T t when t is in R_T. (The sets R_T are sometimes called saturated sets or reducibility candidates.)
Here is the definition of R for the base language:
• R bool t iff t is a closed term of type bool and t halts in a value
• R (T1 T2) t iff t is a closed term of type T1 T2 and t halts in a value and for any term s such that R T1 s, we have R T2 (t s).
This definition gives us the strengthened induction hypothesis that we need. Our primary goal is to show that all programs —-i.e., all closed terms of base type—-halt. But closed terms of base type can contain subterms of functional type, so we need to know something about these as well. Moreover, it is not enough to know that these subterms halt, because the application of a normalized function to a normalized argument involves a substitution, which may enable more evaluation steps. So we need a stronger condition for terms of functional type: not only should they halt themselves, but, when applied to halting arguments, they should yield halting results.
The form of R is characteristic of the logical relations proof technique. (Since we are just dealing with unary relations here, we could perhaps more properly say logical predicates.) If we want to prove some property P of all closed terms of type A, we proceed by proving, by induction on types, that all terms of type A possess property P, all terms of type AA preserve property P, all terms of type (AA)->(AA) preserve the property of preserving property P, and so on. We do this by defining a family of predicates, indexed by types. For the base type A, the predicate is just P. For functional types, it says that the function should map values satisfying the predicate at the input type to values satisfying the predicate at the output type.
When we come to formalize the definition of R in Coq, we hit a problem. The most obvious formulation would be as a parameterized Inductive proposition like this:
Inductive R : ty  tm  Prop :=
R_bool :  b thas_type empty t ty_Bool
halts t
R ty_Bool t
R_arrow :  T1 T2 thas_type empty t (ty_arrow T1 T2
halts t
( sR T1 s  R T2 (tm_app t s))
R (ty_arrow T1 T2t.
Unfortunately, Coq rejects this definition because it violates the strict positivity requirement for inductive definitions, which says that the type being defined must not occur to the left of an arrow in the type of a constructor argument. Here, it is the third argument to R_arrow, namely ( s, R T1 s R TS (tm_app t s)), and specifically the R T1 s part, that violates this rule. (The outermost arrows separating the constructor arguments don't count when applying this rule; otherwise we could never have genuinely inductive predicates at all!) The reason for the rule is that types defined with non-positive recursion can be used to build non-terminating functions, which as we know would be a disaster for Coq's logical soundness. Even though the relation we want in this case might be perfectly innocent, Coq still rejects it because it fails the positivity test.
Fortunately, it turns out that we can define R using a Fixpoint:

Fixpoint R (T:ty) (t:tm) {struct T} : Prop :=
has_type empty t T halts t
(match T with
| ty_Bool => True
| ty_arrow T1 T2 => ( s, R T1 s R T2 (tm_app t s))
(* FILL IN HERE *)
| ty_prod T1 T2 => False (* ... and delete this line *)
end).

As immediate consequences of this definition, we have that every element of every set R_T halts in a value and is closed with type t :

Lemma R_halts : {T} {t}, R T t halts t.
Proof.
intros. destruct T; unfold R in H; inversion H; inversion H1; assumption.
Qed.

Lemma R_typable_empty : {T} {t}, R T t has_type empty t T.
Proof.
intros. destruct T; unfold R in H; inversion H; inversion H1; assumption.
Qed.

Now we proceed to show the main result, which is that every well-typed term of type T is an element of R_T. Together with R_halts, that will show that every well-typed term halts in a value.

## Membership in R_T is invariant under evaluation

We start with a preliminary lemma that shows a kind of strong preservation property, namely that membership in R_T is invariant under evaluation. We will need this property in both directions, i.e. both to show that a term in R_T stays in R_T when it takes a forward step, and to show that any term that ends up in R_T after a step must have been in R_T to begin with.
First of all, an easy preliminary lemma. Note that in the forward direction the proof depends on the fact that our language is determinstic. This lemma might still be true for non-deterministic languages, but the proof would be harder!

Lemma step_preserves_halting : t t', (t t') (halts t halts t').
Proof.
intros t t' ST. unfold halts.
split.
Case "".
intros [t'' [STM V]].
inversion STM; subst.
apply ex_falso_quodlibet. apply value__normal in V. unfold normal_form in V. apply V. t'. auto.
rewrite (step_deterministic _ _ _ ST H). t''. split; assumption.
Case "".
intros [t'0 [STM V]].
t'0. split; eauto.
Qed.

Now the main lemma, which comes in two parts, one for each direction. Each proceeds by induction on the structure of the type T. In fact, this is where we make fundamental use of the finiteness of types.
One requirement for staying in R_T is to stay in type T. In the forward direction, we get this from ordinary type Preservation.

Lemma step_preserves_R : T t t', (t t') R T t R T t'.
Proof.
induction T; intros t t' E Rt; unfold R; fold R; unfold R in Rt; fold R in Rt;
destruct Rt as [typable_empty_t [halts_t RRt]].
(* ty_Bool *)
split. eapply preservation; eauto.
split. apply (step_preserves_halting _ _ E); eauto.
auto.
(* ty_arrow *)
split. eapply preservation; eauto.
split. apply (step_preserves_halting _ _ E); eauto.
intros.
eapply IHT2.
apply ST_App1. apply E.
apply RRt; auto.
(* FILL IN HERE *) Admitted.

The generalization to multiple steps is trivial:

Lemma stepmany_preserves_R : T t t',
(t ⇒* t') R T t R T t'.
Proof.
intros T t t' STM; induction STM; intros.
assumption.
apply IHSTM. eapply step_preserves_R. apply H. assumption.
Qed.

In the reverse direction, we must add the fact that t has type T before stepping as an additional hypothesis.

Lemma step_preserves_R' : T t t',
has_type empty t T (t t') R T t' R T t.
Proof.
(* FILL IN HERE *) Admitted.

Lemma stepmany_preserves_R' : T t t',
has_type empty t T (t ⇒* t') R T t' R T t.
Proof.
intros T t t' HT STM.
induction STM; intros.
assumption.
eapply step_preserves_R'. assumption. apply H. apply IHSTM.
eapply preservation; eauto. auto.
Qed.

## Closed instances of terms of type T belong to R_T

Now we proceed to show that every term of type T belongs to R_T. Here, the induction will be on typing derivations (it would be surprising to see a proof about well-typed terms that did not somewhere involve induction on typing derivations!). The only technical difficulty here is in dealing with the abstraction case. Since we are arguing by induction, the demonstration that a term tm_abs x T1 t2 belongs to R_(T1T2) should involve applying the induction hypothesis to show that t2 belongs to R_(T2). But R_(T2) is defined to be a set of closed terms, while t2 may contain x free, so this does not make sense.
This problem is resolved by using a standard trick to suitably generalize the induction hypothesis: instead of proving a statement involving a closed term, we generalize it to cover all closed instances of an open term t. Informally, the statement of the lemma will look like this:
If x1:T1,..xn:Tn t : T and v1,...,vn are values such that R T1 v1, R T2 v2, ..., R Tn vn, then R T ([v1/x1][v2/x2]...[vn/xn]t).
The proof will proceed by induction on the typing derivation x1:T1,..xn:Tn t : T; the most interesting case will be the one for abstraction.

### Multisubstitutions, multi-extensions, and instantiations

However, before we can proceed to formalize the statement and proof of the lemma, we'll need to build some (rather tedious) machinery to deal with the fact that we are performing multiple substitutions on term t and multiple extensions of the typing context. In particular, we must be precise about the order in which the substitutions occur and how they act on each other. Often these details are simply elided in informal paper proofs, but of course Coq won't let us do that. Since here we are substituting closed terms, we don't need to worry about how one substitution might affect the term put in place by another. But we still do need to worry about the order of substitutions, because it is quite possible for the same identifier to appear multiple times among the x1,...xn with different associated vi and Ti.
To make everything precise, we will assume that environments are extended from left to right, and multiple substitutions are performed from right to left. To see that this is consistent, suppose we have an environment written as ...,y:bool,...,y:nat,... and a corresponding term substitution written as ...[(tm_bool true)/y]...[(tm_nat 3)/y]...t. Since environments are extended from left to right, the binding y:nat hides the binding y:bool; since substitutions are performed right to left, we do the substitution (tm_nat 3)/y first, so that the substitution (tm_bool true)/y has no effect. Substitution thus correctly preserves the type of the term.
With these points in mind, the following definitions should make sense.
A multisubstitution is the result of applying a list of substitutions, which we call an environment.

Definition env := list (id * tm).

Fixpoint msubst (ss:env) (t:tm) {struct ss} : tm :=
match ss with
| nil => t
| ((x,s)::ss') => msubst ss' (subst x s t)
end.

We need similar machinery to talk about repeated extension of a typing context using a list of (identifier, type) pairs, which we call a type assignment.

Definition tass := list (id * ty).

Fixpoint mextend (Γ : context) (xts : tass) :=
match xts with
| nil => Γ
| ((x,v)::xts') => extend (mextend Γ xts') x v
end.

We will need some simple operations that work uniformly on environments and type assigments

Fixpoint lookup {X:Set} (k : id) (l : list (id * X)) {struct l} : option X :=
match l with
| nil => None
| (j,x) :: l' =>
if beq_id j k then Some x else lookup k l'
end.

Fixpoint drop {X:Set} (n:id) (nxs:list (id * X)) {struct nxs} : list (id * X) :=
match nxs with
| nil => nil
| ((n',x)::nxs') => if beq_id n' n then drop n nxs' else (n',x)::(drop n nxs')
end.

An instantiation combines a type assignment and a value environment with the same domains, where corresponding elements are in R

Inductive instantiation : tass env Prop :=
| V_nil : instantiation nil nil
| V_cons : x T v c e, value v R T v instantiation c e instantiation ((x,T)::c) ((x,v)::e).

We now proceed to prove various properties of these definitions.

### More Substitution Facts

First we need some additional lemmas on (ordinary) substitution.

Lemma vacuous_substitution : t x,
~ appears_free_in x t
t', subst x t' t = t.
Proof with eauto.
(* FILL IN HERE *) Admitted.

Lemma subst_closed: t,
closed t
x t', subst x t' t = t.
Proof.
intros. apply vacuous_substitution. apply H. Qed.

Lemma subst_not_afi : t x v, closed v ~ appears_free_in x (subst x v t).
Proof with eauto. (* rather slow this way *)
unfold closed, not.
tm_cases (induction t) Case; intros x v P A; simpl in A.
Case "tm_var".
remember (beq_id x i) as e; destruct e...
inversion A; subst. rewrite beq_id_refl in Heqe; inversion Heqe.
Case "tm_app".
inversion A; subst...
Case "tm_abs".
remember (beq_id x i) as e; destruct e...
apply beq_id_eq in Heqe; subst. inversion A; subst...
inversion A; subst...
Case "tm_pair".
inversion A; subst...
Case "tm_fst".
inversion A; subst...
Case "tm_snd".
inversion A; subst...
Case "tm_true".
inversion A.
Case "tm_false".
inversion A.
Case "tm_if".
inversion A; subst...
Qed.

Lemma duplicate_subst : t' x t v,
closed v subst x t (subst x v t') = subst x v t'.
Proof.
intros. eapply vacuous_substitution. apply subst_not_afi. auto.
Qed.

Lemma swap_subst : t x x1 v v1, x <> x1 closed v closed v1
subst x1 v1 (subst x v t) = subst x v (subst x1 v1 t).
Proof with eauto.
tm_cases (induction t) Case; intros; simpl.
Case "tm_var".
remember (beq_id x i) as e; destruct e; remember (beq_id x1 i) as e; destruct e.
apply beq_id_eq in Heqe. apply beq_id_eq in Heqe0. subst.
apply ex_falso_quodlibet...
apply beq_id_eq in Heqe; subst. simpl.
rewrite beq_id_refl. apply subst_closed...
apply beq_id_eq in Heqe0; subst. simpl.
rewrite beq_id_refl. rewrite subst_closed...
simpl. rewrite Heqe. rewrite Heqe0...
(* FILL IN HERE *) Admitted.

### Properties of multi-substitutions

Lemma msubst_closed: t, closed t ss, msubst ss t = t.
Proof.
induction ss.
reflexivity.
destruct a. simpl. rewrite subst_closed; assumption.
Qed.

Closed environments are those that contain only closed terms.

Fixpoint closed_env (env:env) {struct env} :=
match env with
| nil => True
| (x,t)::env' => closed t closed_env env'
end.

Next come a series of lemmas charcterizing how msubst of closed terms distributes over subst and over each term form

Lemma subst_msubst: env x v t, closed v closed_env env
msubst env (subst x v t) = subst x v (msubst (drop x env) t).
Proof.
induction env0; intros.
auto.
destruct a. simpl.
inversion H0. fold closed_env in H2.
remember (beq_id i x) as e; destruct e.
apply beq_id_eq in Heqe; subst.
rewrite duplicate_subst; auto.
symmetry in Heqe. apply beq_id_false_not_eq in Heqe.
simpl. rewrite swap_subst; eauto.
Qed.

Lemma msubst_var: ss x, closed_env ss
msubst ss (tm_var x) =
match lookup x ss with
| Some t => t
| None => tm_var x
end.
Proof.
induction ss; intros.
reflexivity.
destruct a.
simpl. destruct (beq_id i x).
apply msubst_closed. inversion H; auto.
apply IHss. inversion H; auto.
Qed.

Lemma msubst_abs: ss x T t,
msubst ss (tm_abs x T t) = tm_abs x T (msubst (drop x ss) t).
Proof.
induction ss; intros.
reflexivity.
destruct a.
simpl. destruct (beq_id i x); simpl; auto.
Qed.

Lemma msubst_app : ss t1 t2, msubst ss (tm_app t1 t2) = tm_app (msubst ss t1) (msubst ss t2).
Proof.
induction ss; intros.
reflexivity.
destruct a.
simpl. rewrite IHss. auto.
Qed.

You'll need similar functions for the other term constructors.

(* FILL IN HERE *)

### Properties of multi-extensions

We need to connect the behavior of type assignments with that of their corresponding contexts.

Lemma mextend_lookup : (c : tass) (x:id), lookup x c = (mextend empty c) x.
Proof.
induction c; intros.
auto.
destruct a. unfold lookup, mextend, extend. destruct (beq_id i x); auto.
Qed.

Lemma mextend_drop : (c: tass) Γ x x',
mextend Γ (drop x c) x' = if beq_id x x' then Γ x' else mextend Γ c x'.
induction c; intros.
destruct (beq_id x x'); auto.
destruct a. simpl.
remember (beq_id i x) as e; destruct e.
apply beq_id_eq in Heqe; subst. rewrite IHc.
remember (beq_id x x') as e; destruct e. auto. unfold extend. rewrite Heqe. auto.
simpl. unfold extend. remember (beq_id i x') as e; destruct e.
apply beq_id_eq in Heqe0; subst.
remember (beq_id x x') as e; destruct e.
apply beq_id_eq in Heqe0; subst. rewrite beq_id_refl in Heqe. inversion Heqe.
auto.
auto.
Qed.

### Properties of Instantiations

These are strightforward.

Lemma instantiation_domains_match: {c} {e},
instantiation c e {x} {T}, lookup x c = Some T t, lookup x e = Some t.
Proof.
intros c e V. induction V; intros x0 T0 C.
solve by inversion .
simpl in *.
destruct (beq_id x x0); eauto.
Qed.

Lemma instantiation_env_closed : c e, instantiation c e closed_env e.
Proof.
intros c e V; induction V; intros.
econstructor.
unfold closed_env. fold closed_env.
split. eapply typable_empty__closed. eapply R_typable_empty. eauto.
auto.
Qed.

Lemma instantiation_R : c e, instantiation c e
x t T, lookup x c = Some T
lookup x e = Some t R T t.
Proof.
intros c e V. induction V; intros x' t' T' G E.
solve by inversion.
unfold lookup in *. destruct (beq_id x x').
inversion G; inversion E; subst. auto.
eauto.
Qed.

Lemma instantiation_drop : c env,
instantiation c env x, instantiation (drop x c) (drop x env).
Proof.
intros c e V. induction V.
intros. simpl. constructor.
intros. unfold drop. destruct (beq_id x x0); auto. constructor; eauto.
Qed.

### Congruence lemmas on stepmany

We'll need just a few of these; add them as the demand arises.

Lemma stepmany_App2 : v t t',
value v (t ⇒* t') (tm_app v t) ⇒* (tm_app v t').
Proof.
intros v t t' V STM. induction STM.
apply rsc_refl.
eapply rsc_step.
apply ST_App2; eauto. auto.
Qed.

(* FILL IN HERE *)

### The R Lemma.

We finally put everything together.
The key lemma about preservation of typing under substitution can be lifted to multi-substitutions:

Lemma msubst_preserves_typing : c e,
instantiation c e
Γ t S, has_type (mextend Γ c) t S
has_type Γ (msubst e t) S.
Proof.
induction 1; intros.
simpl in H. simpl. auto.
simpl in H2. simpl.
apply IHinstantiation.
eapply substitution_preserves_typing; eauto.
apply (R_typable_empty H0).
Qed.

And at long last, the main lemma.

Lemma msubst_R : c env t T,
has_type (mextend empty c) t T instantiation c env R T (msubst env t).
Proof.
intros c env0 t T HT V.
generalize dependent env0.
(* We need to generalize the hypothesis a bit before setting up the induction. *)
remember (mextend empty c) as Γ.
assert ( x, Γ x = lookup x c).
intros. rewrite HeqGamma. rewrite mextend_lookup. auto.
clear HeqGamma.
generalize dependent c.
has_type_cases (induction HT) Case; intros.

Case "T_Var".
rewrite H0 in H. destruct (instantiation_domains_match V H) as [t P].
eapply instantiation_R; eauto.
rewrite msubst_var. rewrite P. auto. eapply instantiation_env_closed; eauto.

Case "T_Abs".
rewrite msubst_abs.
(* We'll need variants of the following fact several times, so its simplest to
establish it just once. *)

assert (WT: has_type empty (tm_abs x T11 (msubst (drop x env0) t12)) (ty_arrow T11 T12)).
eapply T_Abs. eapply msubst_preserves_typing. eapply instantiation_drop; eauto.
eapply context_invariance. apply HT.
intros.
unfold extend. rewrite mextend_drop. remember (beq_id x x0) as e; destruct e. auto.
rewrite H.
clear - c Heqe. induction c.
simpl. rewrite Heqe. auto.
simpl. destruct a. unfold extend. destruct (beq_id i x0); auto.
unfold R. fold R. split.
auto.
split. apply value_halts. apply v_abs.
intros.
destruct (R_halts H0) as [v [P Q]].
pose proof (stepmany_preserves_R _ _ _ P H0).
apply stepmany_preserves_R' with (msubst ((x,v)::env0) t12).
eapply T_App. eauto.
apply R_typable_empty; auto.
eapply rsc_trans. eapply stepmany_App2; eauto.
eapply rsc_R.
simpl. rewrite subst_msubst.
eapply ST_AppAbs; eauto.
eapply typable_empty__closed.
apply (R_typable_empty H1).
eapply instantiation_env_closed; eauto.
eapply (IHHT ((x,T11)::c)).
intros. unfold extend, lookup. destruct (beq_id x x0); auto.
constructor; auto.

Case "T_App".
rewrite msubst_app.
destruct (IHHT1 c H env0 V) as [_ [_ P1]].
pose proof (IHHT2 c H env0 V) as P2. fold R in P1. auto.

(* FILL IN HERE *) Admitted.

### Normalization Theorem

Theorem normalization : t T, has_type empty t T halts t.
Proof.
intros.
replace t with (msubst nil t).
eapply R_halts.
eapply msubst_R; eauto. instantiate (2:= nil). eauto.
eapply V_nil.
auto.
Qed.