(** * StlcProp: Properties of STLC *) Require Export Stlc. Module STLCProp. Import STLC. (** In this chapter, we develop the fundamental theory of the Simply Typed Lambda Calculus -- in particular, the type safety theorem. *) (* ###################################################################### *) (** * Canonical Forms *) Lemma canonical_forms_bool : forall t, empty |- t \in TBool -> value t -> (t = ttrue) \/ (t = tfalse). Proof. intros t HT HVal. inversion HVal; intros; subst; try inversion HT; auto. Qed. Lemma canonical_forms_fun : forall t T1 T2, empty |- t \in (TArrow T1 T2) -> value t -> exists x u, t = tabs x T1 u. Proof. intros t T1 T2 HT HVal. inversion HVal; intros; subst; try inversion HT; subst; auto. exists x0. exists t0. auto. Qed. (* ###################################################################### *) (** * Progress *) (** As before, the _progress_ theorem tells us that closed, well-typed terms are not stuck: either a well-typed term is a value, or it can take an evaluation step. The proof is a relatively straightforward extension of the progress proof we saw in the [Types] chapter. *) Theorem progress : forall t T, empty |- t \in T -> value t \/ exists t', t ==> t'. (** _Proof_: by induction on the derivation of [|- t \in 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 \in T2 -> T] and [|- t2 \in T2]. By the induction hypothesis, either [t1] is a value or it can take an evaluation step. - If [t1] is a value, we now consider [t2], which by the other induction hypothesis must also either be a value or take an evaluation step. - Suppose [t2] is a value. Since [t1] is a value with an arrow type, it must be a lambda abstraction; hence [t1 t2] can take a step by [ST_AppAbs]. - Otherwise, [t2] can take a step, and hence so can [t1 t2] by [ST_App2]. - If [t1] can take a step, then so can [t1 t2] by [ST_App1]. - If the last rule of the derivation was [T_If], then [t = if t1 then t2 else t3], where [t1] has type [Bool]. By the IH, [t1] either is a value or takes a step. - If [t1] is a value, then since it has type [Bool] it must be either [true] or [false]. If it is [true], then [t] steps to [t2]; otherwise it steps to [t3]. - Otherwise, [t1] takes a step, and therefore so does [t] (by [ST_If]). *) Proof with eauto. intros t T Ht. remember (@empty ty) as Gamma. has_type_cases (induction Ht) Case; subst Gamma... 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". assert (exists x0 t0, t1 = tabs x0 T11 t0). eapply canonical_forms_fun; eauto. destruct H1 as [x0 [t0 Heq]]. subst. exists ([x0:=t2]t0)... SSCase "t2 steps". inversion H0 as [t2' Hstp]. exists (tapp t1 t2')... SCase "t1 steps". inversion H as [t1' Hstp]. exists (tapp t1' t2)... Case "T_If". right. destruct IHHt1... SCase "t1 is a value". destruct (canonical_forms_bool t1); subst; eauto. SCase "t1 also steps". inversion H as [t1' Hstp]. exists (tif t1' t2 t3)... Qed. (** **** Exercise: 3 stars, optional (progress_from_term_ind) *) (** Show that progress can also be proved by induction on terms instead of induction on typing derivations. *) Theorem progress' : forall t T, empty |- t \in T -> value t \/ exists t', t ==> t'. Proof. intros t. t_cases (induction t) Case; intros T Ht; auto. (* FILL IN HERE *) Admitted. (** [] *) (* ###################################################################### *) (** * Preservation *) (** The other half of the type soundness property is the preservation of types during reduction. For this, we need to develop some technical machinery for reasoning about variables and substitution. Working from top to bottom (the high-level property we are actually interested in to the lowest-level technical lemmas that are needed by various cases of the more interesting proofs), the story goes like this: - The _preservation theorem_ is proved by induction on a typing derivation, pretty much as we did in the [Types] chapter. The one case that is significantly different is the one for the [ST_AppAbs] rule, which is defined using the substitution operation. To see that this step preserves typing, we need to know that the substitution itself does. So we prove a... - _substitution lemma_, stating that substituting a (closed) term [s] for a variable [x] in a term [t] preserves the type of [t]. The proof goes by induction on the form of [t] and requires looking at all the different cases in the definition of substitition. This time, the tricky cases are the ones for variables and for function abstractions. In both cases, we discover that we need to take a term [s] that has been shown to be well-typed in some context [Gamma] and consider the same term [s] in a slightly different context [Gamma']. For this we prove a... - _context invariance_ lemma, showing that typing is preserved under "inessential changes" to the context [Gamma] -- in particular, changes that do not affect any of the free variables of the term. For this, we need a careful definition of - the _free variables_ of a term -- i.e., the variables occuring in the term that are not in the scope of a function abstraction that binds them. *) (* ###################################################################### *) (** ** Free Occurrences *) (** A variable [x] _appears free in_ a term _t_ if [t] contains some occurrence of [x] that is not under an abstraction labeled [x]. For example: - [y] appears free, but [x] does not, in [\x: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 : forall x, appears_free_in x (tvar x) | afi_app1 : forall x t1 t2, appears_free_in x t1 -> appears_free_in x (tapp t1 t2) | afi_app2 : forall x t1 t2, appears_free_in x t2 -> appears_free_in x (tapp t1 t2) | afi_abs : forall x y T11 t12, y <> x -> appears_free_in x t12 -> appears_free_in x (tabs y T11 t12) | afi_if1 : forall x t1 t2 t3, appears_free_in x t1 -> appears_free_in x (tif t1 t2 t3) | afi_if2 : forall x t1 t2 t3, appears_free_in x t2 -> appears_free_in x (tif t1 t2 t3) | afi_if3 : forall x t1 t2 t3, appears_free_in x t3 -> appears_free_in x (tif t1 t2 t3). Tactic Notation "afi_cases" tactic(first) ident(c) := first; [ Case_aux c "afi_var" | Case_aux c "afi_app1" | Case_aux c "afi_app2" | Case_aux c "afi_abs" | Case_aux c "afi_if1" | Case_aux c "afi_if2" | Case_aux c "afi_if3" ]. Hint Constructors appears_free_in. (** A term in which no variables appear free is said to be _closed_. *) Definition closed (t:tm) := forall x, ~ appears_free_in x t. (* ###################################################################### *) (** ** Substitution *) (** We first need a technical lemma connecting free variables and typing contexts. If a variable [x] appears free in a term [t], and if we know [t] is well typed in context [Gamma], then it must be the case that [Gamma] assigns a type to [x]. *) Lemma free_in_context : forall x t T Gamma, appears_free_in x t -> Gamma |- t \in T -> exists T', Gamma x = Some T'. (** _Proof_: We show, by induction on the proof that [x] appears free in [t], that, for all contexts [Gamma], if [t] is well typed under [Gamma], then [Gamma] 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 [Gamma] we have immediately that [Gamma] 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 [Gamma], we can see from the typing rules that [t1] must also be, and the IH then tells us that [Gamma] 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 [Gamma], we know the subterm of [t] in which [x] appears is well typed under [Gamma] 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 [Gamma], its body [t12] is well typed under [(Gamma, y:T11)], so the IH allows us to conclude that [x] is assigned some type by the extended context [(Gamma, y:T11)]. To conclude that [Gamma] assigns a type to [x], we appeal to lemma [extend_neq], noting that [x] and [y] are different variables. *) Proof. intros x t T Gamma H H0. generalize dependent Gamma. 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. 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, optional (typable_empty__closed) *) Corollary typable_empty__closed : forall t T, empty |- t \in T -> closed t. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** Sometimes, when we have a proof [Gamma |- t : T], we will need to replace [Gamma] 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 [Gamma] to all the variables that appear free in [t]. In fact, this is the only condition that is needed. *) Lemma context_invariance : forall Gamma Gamma' t T, Gamma |- t \in T -> (forall x, appears_free_in x t -> Gamma x = Gamma' x) -> Gamma' |- t \in T. (** _Proof_: By induction on the derivation of [Gamma |- t \in T]. - If the last rule in the derivation was [T_Var], then [t = x] and [Gamma x = T]. By assumption, [Gamma' x = T] as well, and hence [Gamma' |- t \in T] by [T_Var]. - If the last rule was [T_Abs], then [t = \y:T11. t12], with [T = T11 -> T12] and [Gamma, y:T11 |- t12 \in T12]. The induction hypothesis is that for any context [Gamma''], if [Gamma, 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 [Gamma] on the free variables in [t]; we must show [Gamma' |- \y:T11. t12 \in T11 -> T12]. By [T_Abs], it suffices to show that [Gamma', y:T11 |- t12 \in T12]. By the IH (setting [Gamma'' = Gamma', y:T11]), it suffices to show that [Gamma, y:T11] and [Gamma', y:T11] agree on all the variables that appear free in [t12]. Any variable occurring free in [t12] must either be [y], or some other variable. [Gamma, y:T11] and [Gamma', y:T11] clearly agree on [y]. Otherwise, we note that any variable other than [y] which occurs free in [t12] also occurs free in [t = \y:T11. t12], and by assumption [Gamma] and [Gamma'] agree on all such variables, and hence so do [Gamma, y:T11] and [Gamma', y:T11]. - If the last rule was [T_App], then [t = t1 t2], with [Gamma |- t1 \in T2 -> T] and [Gamma |- t2 \in T2]. One induction hypothesis states that for all contexts [Gamma'], if [Gamma'] agrees with [Gamma] 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 [Gamma] 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 [Gamma]. 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 x1 Hafi. (* the only tricky step... the [Gamma'] we use to instantiate is [extend Gamma x T11] *) unfold extend. destruct (eq_id_dec x0 x1)... Case "T_App". apply T_App with T11... Qed. (** Now we come to the conceptual heart of the proof that reduction preserves types -- namely, the observation that _substitution_ preserves types. Formally, the so-called _Substitution Lemma_ says this: suppose we have a term [t] with a free variable [x], and suppose we've been able to assign a type [T] to [t] under the assumption that [x] has some type [U]. Also, suppose that we have some other term [v] and that we've shown that [v] has type [U]. Then, since [v] satisfies the assumption we made about [x] when typing [t], we should be able to substitute [v] for each of the occurrences of [x] in [t] and obtain a new term that still has type [T]. *) (** _Lemma_: If [Gamma,x:U |- t \in T] and [|- v \in U], then [Gamma |- [x:=v]t \in T]. *) Lemma substitution_preserves_typing : forall Gamma x U t v T, extend Gamma x U |- t \in T -> empty |- v \in U -> Gamma |- [x:=v]t \in T. (** One technical subtlety in the statement of the lemma is that we assign [v] the type [U] in the _empty_ context -- in other words, we assume [v] is closed. This assumption considerably simplifies the [T_Abs] case of the proof (compared to assuming [Gamma |- v \in 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 [Gamma], if [Gamma,x:U |- t \in T] and [|- v \in U], then [Gamma |- [x:=v]t \in 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 [Gamma, x:U |- x \in T] we conclude that [U = T]. We must show that [[x:=v]x = v] has type [T] under [Gamma], 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 [Gamma, x:U] as under [Gamma]. - If [t] is an abstraction [\y:T11. t12], then the IH tells us, for all [Gamma'] and [T'], that if [Gamma',x:U |- t12 \in T'] and [|- v \in U], then [Gamma' |- [x:=v]t12 \in T']. The substitution in the conclusion behaves differently, depending on whether [x] and [y] are the same variable name. First, suppose [x = y]. Then, by the definition of substitution, [[x:=v]t = t], so we just need to show [Gamma |- t \in T]. But we know [Gamma,x:U |- t : T], and since the variable [y] does not appear free in [\y:T11. t12], the context invariance lemma yields [Gamma |- t \in T]. Second, suppose [x <> y]. We know [Gamma,x:U,y:T11 |- t12 \in T12] by inversion of the typing relation, and [Gamma,y:T11,x:U |- t12 \in T12] follows from this by the context invariance lemma, so the IH applies, giving us [Gamma,y:T11 |- [x:=v]t12 \in T12]. By [T_Abs], [Gamma |- \y:T11. [x:=v]t12 \in T11->T12], and by the definition of substitution (noting that [x <> y]), [Gamma |- \y:T11. [x:=v]t12 \in T11->T12] as required. - If [t] is an application [t1 t2], the result follows straightforwardly from the definition of substitution and the induction hypotheses. - The remaining cases are similar to the application case. Another technical note: This proof is a rare case where an induction on terms, rather than typing derivations, yields a simpler argument. The reason for this is that the assumption [extend Gamma x U |- t \in T] is not completely generic, in the sense that one of the "slots" in the typing relation -- namely the context -- is not just a variable, and this means that Coq's native induction tactic does not give us the induction hypothesis that we want. It is possible to work around this, but the needed generalization is a little tricky. The term [t], on the other hand, _is_ completely generic. *) Proof with eauto. intros Gamma x U t v T Ht Ht'. generalize dependent Gamma. generalize dependent T. t_cases (induction t) Case; intros T Gamma H; (* in each case, we'll want to get at the derivation of H *) inversion H; subst; simpl... Case "tvar". rename i into y. destruct (eq_id_dec x y). SCase "x=y". subst. rewrite extend_eq in H2. inversion H2; subst. clear H2. eapply context_invariance... intros x Hcontra. destruct (free_in_context _ _ T empty Hcontra) as [T' HT']... inversion HT'. SCase "x<>y". apply T_Var. rewrite extend_neq in H2... Case "tabs". rename i into y. apply T_Abs. destruct (eq_id_dec x y). SCase "x=y". eapply context_invariance... subst. intros x Hafi. unfold extend. destruct (eq_id_dec y x)... SCase "x<>y". apply IHt. eapply context_invariance... intros z Hafi. unfold extend. destruct (eq_id_dec y z)... subst. rewrite neq_id... Qed. (** The substitution lemma can be viewed as a kind of "commutation" property. Intuitively, it says that substitution and typing can be done in either order: we can either assign types to the terms [t] and [v] separately (under suitable contexts) and then combine them using substitution, or we can substitute first and then assign a type to [ [x:=v] t ] -- the result is the same either way. *) (* ###################################################################### *) (** ** Main Theorem *) (** We now have the tools we need to prove preservation: if a closed term [t] has type [T], and takes an evaluation step to [t'], then [t'] is also a closed term with type [T]. In other words, the small-step evaluation relation preserves types. *) Theorem preservation : forall t t' T, empty |- t \in T -> t ==> t' -> empty |- t' \in T. (** _Proof_: by induction on the derivation of [|- t \in T]. - We can immediately rule out [T_Var], [T_Abs], [T_True], and [T_False] as the final rules in the derivation, since in each of these cases [t] cannot take a step. - If the last rule in the derivation was [T_App], then [t = t1 t2]. There are three cases to consider, one for each rule that could have been used to show that [t1 t2] takes a step to [t']. - If [t1 t2] takes a step by [ST_App1], with [t1] stepping to [t1'], then by the IH [t1'] has the same type as [t1], and hence [t1' t2] has the same type as [t1 t2]. - The [ST_App2] case is similar. - If [t1 t2] takes a step by [ST_AppAbs], then [t1 = \x:T11.t12] and [t1 t2] steps to [[x:=t2]t12]; the desired result now follows from the fact that substitution preserves types. - If the last rule in the derivation was [T_If], then [t = if t1 then t2 else t3], and there are again three cases depending on how [t] steps. - If [t] steps to [t2] or [t3], the result is immediate, since [t2] and [t3] have the same type as [t]. - Otherwise, [t] steps by [ST_If], and the desired conclusion follows directly from the induction hypothesis. *) Proof with eauto. remember (@empty ty) as Gamma. intros t t' T HT. generalize dependent t'. has_type_cases (induction HT) Case; intros t' HE; subst Gamma; subst; try solve [inversion HE; subst; auto]. Case "T_App". inversion HE; subst... (* Most of the cases are immediate by induction, and [eauto] takes care of them *) SCase "ST_AppAbs". apply substitution_preserves_typing with T11... inversion HT1... Qed. (** **** Exercise: 2 stars (subject_expansion_stlc) *) (** An exercise in the [Types] chapter 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 [empty |- t \in T]? If so, prove it. If not, give a counter-example not involving conditionals. (* FILL IN HERE *) [] *) (* ###################################################################### *) (** * Type Soundness *) (** **** Exercise: 2 stars, optional (type_soundness) *) (** Put progress and preservation together and show that a well-typed term can _never_ reach a stuck state. *) Definition stuck (t:tm) : Prop := (normal_form step) t /\ ~ value t. Corollary soundness : forall t t' T, empty |- t \in T -> t ==>* t' -> ~(stuck t'). Proof. intros t t' T Hhas_type Hmulti. unfold stuck. intros [Hnf Hnot_val]. unfold normal_form in Hnf. induction Hmulti. (* FILL IN HERE *) Admitted. (* ###################################################################### *) (** * Uniqueness of Types *) (** **** Exercise: 3 stars (types_unique) *) (** Another pleasant property of the STLC is that types are unique: a given term (in a given context) has at most one type. *) (** Formalize this statement and prove it. *) (* FILL IN HERE *) (** [] *) (* ###################################################################### *) (** * 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 a new term [zap] with the following reduction rule: --------- (ST_Zap) t ==> zap and the following typing rule: ---------------- (T_Zap) Gamma |- zap : T Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample. - Determinism of [step] - Progress - Preservation [] *) (** **** Exercise: 2 stars (stlc_variation2) *) (** Suppose instead that we add a new term [foo] with the following reduction rules: ----------------- (ST_Foo1) (\x:A. x) ==> foo ------------ (ST_Foo2) foo ==> true Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample. - Determinism of [step] - Progress - Preservation [] *) (** **** Exercise: 2 stars (stlc_variation3) *) (** Suppose instead that we remove the rule [ST_App1] from the [step] relation. Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample. - Determinism of [step] - Progress - Preservation [] *) (** **** Exercise: 2 stars, optional (stlc_variation4) *) (** Suppose instead that we add the following new rule to the reduction relation: ---------------------------------- (ST_FunnyIfTrue) (if true then t1 else t2) ==> true Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample. - Determinism of [step] - Progress - Preservation *) (** **** Exercise: 2 stars, optional (stlc_variation5) *) (** Suppose instead that we add the following new rule to the typing relation: Gamma |- t1 \in Bool->Bool->Bool Gamma |- t2 \in Bool ------------------------------ (T_FunnyApp) Gamma |- t1 t2 \in Bool Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample. - Determinism of [step] - Progress - Preservation *) (** **** Exercise: 2 stars, optional (stlc_variation6) *) (** Suppose instead that we add the following new rule to the typing relation: Gamma |- t1 \in Bool Gamma |- t2 \in Bool --------------------- (T_FunnyApp') Gamma |- t1 t2 \in Bool Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample. - Determinism of [step] - Progress - Preservation *) (** **** Exercise: 2 stars, optional (stlc_variation7) *) (** Suppose we add the following new rule to the typing relation of the STLC: ------------------- (T_FunnyAbs) |- \x:Bool.t \in Bool Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample. - Determinism of [step] - Progress - Preservation [] *) End STLCProp. (* ###################################################################### *) (* ###################################################################### *) (** ** Exercise: STLC with Arithmetic *) (** To see how the STLC might function as the core of a real programming language, let's extend it with a concrete base type of numbers and some constants and primitive operators. *) Module STLCArith. (** To types, we add a base type of natural numbers (and remove booleans, for brevity) *) Inductive ty : Type := | TArrow : ty -> ty -> ty | TNat : ty. (** To terms, we add natural number constants, along with successor, predecessor, multiplication, and zero-testing... *) Inductive tm : Type := | tvar : id -> tm | tapp : tm -> tm -> tm | tabs : id -> ty -> tm -> tm | tnat : nat -> tm | tsucc : tm -> tm | tpred : tm -> tm | tmult : tm -> tm -> tm | tif0 : tm -> tm -> tm -> tm. Tactic Notation "t_cases" tactic(first) ident(c) := first; [ Case_aux c "tvar" | Case_aux c "tapp" | Case_aux c "tabs" | Case_aux c "tnat" | Case_aux c "tsucc" | Case_aux c "tpred" | Case_aux c "tmult" | Case_aux c "tif0" ]. (** **** Exercise: 4 stars (stlc_arith) *) (** Finish formalizing the definition and properties of the STLC extended with arithmetic. Specifically: - Copy the whole development of STLC that we went through above (from the definition of values through the Progress theorem), and paste it into the file at this point. - Extend the definitions of the [subst] operation and the [step] relation to include appropriate clauses for the arithmetic operators. - Extend the proofs of all the properties (up to [soundness]) of the original STLC to deal with the new syntactic forms. Make sure Coq accepts the whole file. *) (* FILL IN HERE *) (** [] *) End STLCArith. (** $Date: 2014-12-31 11:17:56 -0500 (Wed, 31 Dec 2014) $ *)