(** * 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 cannonical_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 cannonical_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 cannonical_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 (cannonical_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: 2013-11-20 13:03:49 -0500 (Wed, 20 Nov 2013) $ *)