(** * StlcProp: Properties of STLC *) Set Warnings "-notation-overridden,-parsing". From PLF Require Import Maps. From PLF Require Import Types. From PLF Require Import Stlc. From PLF Require Import Smallstep. 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 *) (** As we saw for the simple calculus in the [Types] chapter, the first step in establishing basic properties of reduction and types is to identify the possible _canonical forms_ (i.e., well-typed closed values) belonging to each type. For [Bool], these are the boolean values [tru] and [fls]; for arrow types, they are lambda-abstractions. *) Lemma canonical_forms_bool : forall t, empty |- t \in Bool -> value t -> (t = tru) \/ (t = fls). 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 (Arrow T1 T2) -> value t -> exists x u, t = abs x T1 u. Proof. intros t T1 T2 HT HVal. inversion HVal; intros; subst; try inversion HT; subst; auto. exists x0, t0. auto. Qed. (* ################################################################# *) (** * Progress *) (** 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 a reduction step. The proof is a relatively straightforward extension of the progress proof we saw in the [Types] chapter. We give the proof in English first, then the formal version. *) 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_Tru], [T_Fls], and [T_Abs] cases are trivial, since in each of these cases we can see by inspecting the rule that [t] is a value. - If the last rule of the derivation is [T_App], then [t] has the form [t1 t2] for some [t1] and [t2], where [|- t1 \in T2 -> T] and [|- t2 \in T2] for some type [T2]. The induction hypothesis for the first subderivation says that either [t1] is a value or else it can take a reduction step. - If [t1] is a value, then consider [t2], which by the induction hypothesis for the second subderivation must also either be a value or take a 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 is [T_Test], then [t = test t1 then t2 else t3], where [t1] has type [Bool]. The first IH says that [t1] either is a value or takes a step. - If [t1] is a value, then since it has type [Bool] it must be either [tru] or [fls]. If it is [tru], then [t] steps to [t2]; otherwise it steps to [t3]. - Otherwise, [t1] takes a step, and therefore so does [t] (by [ST_Test]). *) Proof with eauto. intros t T Ht. remember (@empty ty) as Gamma. induction Ht; subst Gamma... - (* T_Var *) (* contradictory: variables cannot be typed in an empty context *) inversion H. - (* T_App *) (* [t] = [t1 t2]. Proceed by cases on whether [t1] is a value or steps... *) right. destruct IHHt1... + (* t1 is a value *) destruct IHHt2... * (* t2 is also a value *) eapply canonical_forms_fun in H... destruct H as [x0 [t0 Heq]]; subst. exists ([x0:=t2]t0)... * (* t2 steps *) inversion H0 as [t2' Hstp]. exists (app t1 t2')... + (* t1 steps *) inversion H as [t1' Hstp]. exists (app t1' t2)... - (* T_Test *) right. destruct IHHt1... + (* t1 is a value *) destruct (canonical_forms_bool t1); subst; eauto. + (* t1 also steps *) inversion H as [t1' Hstp]. exists (test t1' t2 t3)... Qed. (** **** Exercise: 3 stars, advanced (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. induction t; 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 part, we'll need to develop some technical machinery for reasoning about variables and substitution. Working from top to bottom (from 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, whose definition uses 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, 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. And finally, for this, we need a careful definition of... - the _free variables_ in a term -- i.e., variables that are used in the term in positions that are _not_ in the scope of an enclosing function abstraction binding a variable of the same name. To make Coq happy, of course, we need to formalize the story in the opposite order... *) (* ================================================================= *) (** ** 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] Formally: *) Inductive appears_free_in : string -> tm -> Prop := | afi_var : forall x, appears_free_in x (var x) | afi_app1 : forall x t1 t2, appears_free_in x t1 -> appears_free_in x (app t1 t2) | afi_app2 : forall x t1 t2, appears_free_in x t2 -> appears_free_in x (app t1 t2) | afi_abs : forall x y T11 t12, y <> x -> appears_free_in x t12 -> appears_free_in x (abs y T11 t12) | afi_test1 : forall x t1 t2 t3, appears_free_in x t1 -> appears_free_in x (test t1 t2 t3) | afi_test2 : forall x t1 t2 t3, appears_free_in x t2 -> appears_free_in x (test t1 t2 t3) | afi_test3 : forall x t1 t2 t3, appears_free_in x t3 -> appears_free_in x (test t1 t2 t3). Hint Constructors appears_free_in. (** The _free variables_ of a term are just the variables that appear free in it. A term with no free variables is said to be _closed_. *) Definition closed (t:tm) := forall x, ~ appears_free_in x t. (** An _open_ term is one that may contain free variables. (I.e., every term is an open term; the closed terms are a subset of the open ones. "Open" precisely means "possibly containing free variables.") *) (** **** Exercise: 1 star, standard, optional (afi) In the space below, write out the rules of the [appears_free_in] relation in informal inference-rule notation. (Use whatever notational conventions you like -- the point of the exercise is just for you to think a bit about the meaning of each rule.) Although this is a rather low-level, technical definition, understanding it is crucial to understanding substitution and its properties, which are really the crux of the lambda-calculus. *) (* FILL IN HERE *) (* Do not modify the following line: *) Definition manual_grade_for_afi : option (nat*string) := None. (** [] *) (* ================================================================= *) (** ** Substitution *) (** To prove that substitution preserves typing, 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 is [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 is [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], and 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 [(y|->T11; Gamma], so the IH allows us to conclude that [x] is assigned some type by the extended context [(y|->T11; Gamma]. To conclude that [Gamma] assigns a type to [x], we appeal to lemma [update_neq], noting that [x] and [y] are different variables. *) Proof. intros x t T Gamma H H0. generalize dependent Gamma. generalize dependent T. induction H; intros; try solve [inversion H0; eauto]. - (* afi_abs *) inversion H1; subst; clear H1. apply IHappears_free_in in H7. rewrite update_neq in H7; assumption. Qed. (** From the [free_in_context] lemma, it immediately follows that any term [t] that is well typed in the empty context is closed (it has no free variables). *) (** **** Exercise: 2 stars, standard, 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 of some typing relation [Gamma |- t \in 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 [y|->T11; Gamma |- t12 \in T12]. The induction hypothesis is that, for any context [Gamma''], if [y|->T11; Gamma] 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 [y|->T11; Gamma' |- t12 \in T12]. By the IH (setting [Gamma'' = y|->T11;Gamma']), it suffices to show that [y|->T11;Gamma] and [y|->T11;Gamma'] agree on all the variables that appear free in [t12]. Any variable occurring free in [t12] must be either [y] or some other variable. [y|->T11; Gamma] and [y|->T11; Gamma'] clearly agree on [y]. Otherwise, note that any variable other than [y] that occurs free in [t12] also occurs free in [t = \y:T11. t12], and by assumption [Gamma] and [Gamma'] agree on all such variables; hence so do [y|->T11; Gamma] and [y|->T11; Gamma']. - 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]. But all free variables in [t1] are also free in [t1 t2], and similarly for [t2]; hence the desired result follows from the induction hypotheses. *) Proof with eauto. intros. generalize dependent Gamma'. induction H; intros; auto. - (* T_Var *) apply T_Var. rewrite <- H0... - (* T_Abs *) apply T_Abs. apply IHhas_type. intros x1 Hafi. (* the only tricky step... the [Gamma'] we use to instantiate is [x|->T11;Gamma] *) unfold update. unfold t_update. destruct (eqb_string x0 x1) eqn: Hx0x1... rewrite eqb_string_false_iff in Hx0x1. auto. - (* 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 assigned 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 can substitute [v] for each of the occurrences of [x] in [t] and obtain a new term that still has type [T]. *) (** _Lemma_: If [x|->U; Gamma |- 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, (x |-> U ; Gamma) |- 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 assume [v] has 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]. 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. _Proof_: We show, by induction on [t], that for all [T] and [Gamma], if [x|->U; Gamma |- 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 [x|->U; Gamma |- 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 [x|->U; Gamma] as under [Gamma]. - If [t] is an abstraction [\y:T11. t12], then the IH tells us, for all [Gamma'] and [T'], that if [x|->U; Gamma' |- 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. 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 [x|->U; Gamma |- t \in T], and, since [y] does not appear free in [\y:T11. t12], the context invariance lemma yields [Gamma |- t \in T]. Second, suppose [x <> y]. We know [y|->T11; x|->U; Gamma |- t12 \in T12] by inversion of the typing relation. Since [x <> y], we have [y|->T11; x|->U; Gamma = x|->U; y|->T11; Gamma]. So the IH applies, giving us [y|->T11; Gamma |- [x:=v]t12 \in T12]. - 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. _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 [x|->U; Gamma |- 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. induction t; intros T Gamma H; (* in each case, we'll want to get at the derivation of H *) inversion H; subst; simpl... - (* var *) rename s into y. destruct (eqb_stringP x y) as [Hxy|Hxy]. + (* x=y *) clear H; subst. rewrite update_eq in H2. inversion H2; subst; clear H2. eapply context_invariance. eassumption. apply typable_empty__closed in Ht'. unfold closed in Ht'. intros. exfalso; eapply Ht'; eassumption. + (* x<>y *) apply T_Var. rewrite update_neq in H2... - (* abs *) rename s into y. rename t into T. apply T_Abs. destruct (eqb_stringP x y) as [Hxy | Hxy]. + (* x=y *) subst. rewrite update_shadow in H5. apply H5. + (* x<>y *) apply IHt. rewrite update_permute... Qed. (* ================================================================= *) (** ** Main Theorem *) (** We now have the tools we need to prove preservation: if a closed term [t] has type [T] and takes a step to [t'], then [t'] is also a closed term with type [T]. In other words, the small-step reduction 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_Tru], and [T_Fls] as final rules in the derivation, since in each of these cases [t] cannot take a step. - If the last rule in the derivation is [T_App], then [t = t1 t2], and there are subderivations showing that [|- t1 \in T11->T] and [|- t2 \in T11] plus two induction hypotheses: (1) [t1 --> t1'] implies [|- t1' \in T11->T] and (2) [t2 --> t2'] implies [|- t2' \in T11]. There are now three subcases to consider, one for each rule that could be 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 first IH, [t1'] has the same type as [t1] ([|- t1 \in T11->T]), and hence by [T_App] [t1' t2] has type [T]. - The [ST_App2] case is similar, using the second IH. - 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 substitution lemma. - If the last rule in the derivation is [T_Test], then [t = test t1 then t2 else t3], with [|- t1 \in Bool], [|- t2 \in T], and [|- t3 \in T], and with three induction hypotheses: (1) [t1 --> t1'] implies [|- t1' \in Bool], (2) [t2 --> t2'] implies [|- t2' \in T], and (3) [t3 --> t3'] implies [|- t3' \in T]. There are again three subcases to consider, depending on how [t] steps. - If [t] steps to [t2] or [t3] by [ST_TestTru] or [ST_TestFalse], the result is immediate, since [t2] and [t3] have the same type as [t]. - Otherwise, [t] steps by [ST_Test], and the desired conclusion follows directly from the first induction hypothesis. *) Proof with eauto. remember (@empty ty) as Gamma. intros t t' T HT. generalize dependent t'. induction HT; intros t' HE; subst Gamma; subst; try solve [inversion HE; subst; auto]. - (* T_App *) inversion HE; subst... (* Most of the cases are immediate by induction, and [eauto] takes care of them *) + (* ST_AppAbs *) apply substitution_preserves_typing with T11... inversion HT1... Qed. (** **** Exercise: 2 stars, standard, recommended (subject_expansion_stlc) An exercise in the [Types] chapter asked about the _subject expansion_ property for the simple language of arithmetic and boolean expressions. This property did not hold for that language, and it also fails for STLC. That is, it is not always the case that, if [t --> t'] and [has_type t' T], then [empty |- t \in T]. Show this by giving a counter-example that does _not involve conditionals_. You can state your counterexample informally in words, with a brief explanation. *) (* FILL IN HERE *) (* Do not modify the following line: *) Definition manual_grade_for_subject_expansion_stlc : option (nat*string) := None. (** [] *) (* ################################################################# *) (** * Type Soundness *) (** **** Exercise: 2 stars, standard, 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, standard (unique_types) Another nice property of the STLC is that types are unique: a given term (in a given context) has at most one type. *) Theorem unique_types : forall Gamma e T T', Gamma |- e \in T -> Gamma |- e \in T' -> T = T'. Proof. (* FILL IN HERE *) Admitted. (** [] *) (* ################################################################# *) (** * Additional Exercises *) (** **** Exercise: 1 star, standard, optional (progress_preservation_statement) Without peeking at their statements above, write down the progress and preservation theorems for the simply typed lambda-calculus (as Coq theorems). You can write [Admitted] for the proofs. *) (* FILL IN HERE *) (* Do not modify the following line: *) Definition manual_grade_for_progress_preservation_statement : option (nat*string) := None. (** [] *) (** **** Exercise: 2 stars, standard, optional (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 \in T Which of the following properties of the STLC remain true in the presence of these rules? For each property, write either "remains true" or "becomes false." If a property becomes false, give a counterexample. - Determinism of [step] (* FILL IN HERE *) - Progress (* FILL IN HERE *) - Preservation (* FILL IN HERE *) *) (* Do not modify the following line: *) Definition manual_grade_for_stlc_variation1 : option (nat*string) := None. (** [] *) (** **** Exercise: 2 stars, standard, optional (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 --> tru 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] (* FILL IN HERE *) - Progress (* FILL IN HERE *) - Preservation (* FILL IN HERE *) *) (* Do not modify the following line: *) Definition manual_grade_for_stlc_variation2 : option (nat*string) := None. (** [] *) (** **** Exercise: 2 stars, standard, optional (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] (* FILL IN HERE *) - Progress (* FILL IN HERE *) - Preservation (* FILL IN HERE *) *) (* Do not modify the following line: *) Definition manual_grade_for_stlc_variation3 : option (nat*string) := None. (** [] *) (** **** Exercise: 2 stars, standard, optional (stlc_variation4) Suppose instead that we add the following new rule to the reduction relation: ---------------------------------- (ST_FunnyTestTru) (test tru then t1 else t2) --> tru 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] (* FILL IN HERE *) - Progress (* FILL IN HERE *) - Preservation (* FILL IN HERE *) *) (** [] *) (** **** Exercise: 2 stars, standard, 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] (* FILL IN HERE *) - Progress (* FILL IN HERE *) - Preservation (* FILL IN HERE *) *) (** [] *) (** **** Exercise: 2 stars, standard, 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] (* FILL IN HERE *) - Progress (* FILL IN HERE *) - Preservation (* FILL IN HERE *) *) (** [] *) (** **** Exercise: 2 stars, standard, 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] (* FILL IN HERE *) - Progress (* FILL IN HERE *) - Preservation (* FILL IN HERE *) *) (** [] *) 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. Import STLC. (** To types, we add a base type of natural numbers (and remove booleans, for brevity). *) Inductive ty : Type := | Arrow : ty -> ty -> ty | Nat : ty. (** To terms, we add natural number constants, along with successor, predecessor, multiplication, and zero-testing. *) Inductive tm : Type := | var : string -> tm | app : tm -> tm -> tm | abs : string -> ty -> tm -> tm | const : nat -> tm | scc : tm -> tm | prd : tm -> tm | mlt : tm -> tm -> tm | test0 : tm -> tm -> tm -> tm. (** **** Exercise: 5 stars, standard, optional (stlc_arith) Finish formalizing the definition and properties of the STLC extended with arithmetic. This is a longer exercise. Specifically: 1. Copy the core definitions for STLC that we went through, as well as the key lemmas and theorems, and paste them into the file at this point. Do not copy examples, exercises, etc. (In particular, make sure you don't copy any of the [] comments at the end of exercises, to avoid confusing the autograder.) You should copy over five definitions: - Fixpoint susbt - Inductive value - Inductive step - Inductive has_type - Inductive appears_free_in And five theorems, with their proofs: - Lemma context_invariance - Lemma free_in_context - Lemma substitution_preserves_typing - Theorem preservation - Theorem progress It will be helpful to also copy over "Reserved Notation", "Notation", and "Hint Constructors" for these things. 2. Edit and extend the five definitions (subst, value, step, has_type, and appears_free_in) so they are appropriate for the new STLC extended with arithmetic. 3. Extend the proofs of all the five properties of the original STLC to deal with the new syntactic forms. Make sure Coq accepts the whole file. *) (* FILL IN HERE *) (* Do not modify the following line: *) Definition manual_grade_for_stlc_arith : option (nat*string) := None. (** [] *) End STLCArith. (* Fri Nov 22 20:32:31 EST 2019 *)