(** * StlcProp: Properties of STLC *)
Require Import Maps.
Require Import Types.
Require Import Stlc.
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 [ttrue] and [tfalse]. For arrow types, the canonical forms
are lambda-abstractions. *)
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 *)
(** 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'll 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_True], [T_False], 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]. By the induction
hypothesis, either [t1] is a value or it can take a reduction
step.
- If [t1] is a value, then consider [t2], which by the other
induction hypothesis 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_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.
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 *)
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)...
* (* t2 steps *)
inversion H0 as [t2' Hstp]. exists (tapp t1 t2')...
+ (* t1 steps *)
inversion H as [t1' Hstp]. exists (tapp t1' t2)...
- (* T_If *)
right. destruct IHHt1...
+ (* t1 is a value *)
destruct (canonical_forms_bool t1); subst; eauto.
+ (* t1 also steps *)
inversion H as [t1' Hstp]. exists (tif 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_ of a term -- i.e., those variables
mentioned in a term and not in the scope of an enclosing
function abstraction binding a variable of the same name.
To make Coq happy, 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 : 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).
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 is not closed (or not known to be
closed). *)
(** **** Exercise: 1 starM (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 *)
(** [] *)
(* ================================================================= *)
(** ** 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 [(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
[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.
apply IHappears_free_in in H7.
rewrite update_neq in H7; assumption.
Qed.
(** Next, we'll need the fact that any term [t] that is well typed in
the empty context is closed (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 be either [y] or
some other variable. [Gamma, y:T11] and [Gamma', y:T11]
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 [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]. 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 [update Gamma x T11] *)
unfold update. unfold t_update. destruct (beq_id x0 x1) eqn: Hx0x1...
rewrite beq_id_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 [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,
update 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].
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 [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.
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 [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, from which
[Gamma,y:T11,x:U |- t12 \in T12] follows 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.
_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
[update 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.
induction t; intros T Gamma H;
(* in each case, we'll want to get at the derivation of H *)
inversion H; subst; simpl...
- (* tvar *)
rename i into y. destruct (beq_idP x y) as [Hxy|Hxy].
+ (* x=y *)
subst.
rewrite update_eq in H2.
inversion H2; subst.
eapply context_invariance. eassumption.
apply typable_empty__closed in Ht'. unfold closed in Ht'.
intros. apply (Ht' x0) in H0. inversion H0.
+ (* x<>y *)
apply T_Var. rewrite update_neq in H2...
- (* tabs *)
rename i into y. rename t into T. apply T_Abs.
destruct (beq_idP x y) as [Hxy | Hxy].
+ (* x=y *)
subst. rewrite update_shadow in H5. apply H5.
+ (* x<>y *)
apply IHt. eapply context_invariance...
intros z Hafi. unfold update, t_update.
destruct (beq_idP y z) as [Hyz | Hyz]; subst; trivial.
rewrite <- beq_id_false_iff in Hxy.
rewrite Hxy...
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_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 is [T_App], then [t = t1
t2]. There are three cases 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 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 is [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'.
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, recommendedM (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 starsM (types_unique) *)
(** Another nice 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 starM (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). *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 2 starsM (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 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 *)
[]
*)
(** **** Exercise: 2 starsM (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]
(* FILL IN HERE *)
- Progress
(* FILL IN HERE *)
- Preservation
(* FILL IN HERE *)
[]
*)
(** **** Exercise: 2 starsM (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 *)
[]
*)
(** **** 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]
(* FILL IN HERE *)
- Progress
(* FILL IN HERE *)
- Preservation
(* FILL IN HERE *)
[]
*)
(** **** 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]
(* FILL IN HERE *)
- Progress
(* FILL IN HERE *)
- Preservation
(* FILL IN HERE *)
[]
*)
(** **** 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]
(* FILL IN HERE *)
- Progress
(* FILL IN HERE *)
- Preservation
(* FILL IN HERE *)
[]
*)
(** **** 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]
(* 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 :=
| 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.
(** **** Exercise: 4 starsM (stlc_arith) *)
(** Finish formalizing the definition and properties of the STLC
extended with arithmetic. Specifically:
- Copy the core definitions and theorems for STLC that we went
through above (from the definition of values through the
Preservation theorem, inclusive), and paste it 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.)
- 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 [preservation])
of the original STLC to deal with the new syntactic forms. Make
sure Coq accepts the whole file. *)
(* FILL IN HERE *)
(** [] *)
End STLCArith.
(** $Date: 2016-12-20 12:03:19 -0500 (Tue, 20 Dec 2016) $ *)