(** * Types: Type Systems *)
(* $Date: 2011-06-03 13:58:55 -0400 (Fri, 03 Jun 2011) $ *)
Require Export Ch10_Smallstep.
(** Our next topic, a large one, is _type systems_ -- static program
analyses that classify expressions according to the "shapes" of
their results. We'll begin with a typed version of a very simple
language with just booleans and numbers, to introduce the basic
ideas of types, typing rules, and the fundamental theorems about
type systems: _type preservation_ and _progress_. Then we'll move
on to the _simply typed lambda-calculus_, which lives at the core
of every modern functional programming language (including
Coq). *)
(* ###################################################################### *)
(** * More Automation *)
(** Before we start, let's spend a little time learning to use
some of Coq's more powerful automation features... *)
(* ###################################################################### *)
(** ** The [auto] and [eauto] Tactics *)
(** The [auto] tactic solves goals that are solvable by any combination of
- [intros],
- [apply] (with a local hypothesis, by default), and
- [reflexivity].
The [eauto] tactic works just like [auto], except that it uses
[eapply] instead of [apply]. *)
(** Using [auto] is always "safe" in the sense that it will never fail
and will never change the proof state: either it completely solves
the current goal, or it does nothing.
Here is a contrived example: *)
Lemma auto_example_1 : forall P Q R S T U : Prop,
(P -> Q) ->
(P -> R) ->
(T -> R) ->
(S -> T -> U) ->
((P->Q) -> (P->S)) ->
T ->
P ->
U.
Proof. auto. Qed.
(** When searching for potential proofs of the current goal, [auto]
and [eauto] consider the hypotheses in the current context
together with a _hint database_ of other lemmas and constructors.
Some of the lemmas and constructors we've already seen -- e.g.,
[conj], [or_introl], and [or_intror] -- are installed in this hint
database by default. *)
Lemma auto_example_2 : forall P Q R : Prop,
Q ->
(Q -> R) ->
P \/ (Q /\ R).
Proof.
auto. Qed.
(** We can extend the hint database just for the purposes of one
application of [auto] or [eauto] by writing [auto using ...].
E.g., if [conj], [or_introl], and [or_intror] had _not_ already
been in the hint database, we could have done this instead: *)
Lemma auto_example_2a : forall P Q R : Prop,
Q ->
(Q -> R) ->
P \/ (Q /\ R).
Proof.
auto using conj, or_introl, or_intror. Qed.
(** Of course, in any given development there will also be some of our
own specific constructors and lemmas that are used very often in
proofs. We can add these to the global hint database by writing
Hint Resolve T.
at the top level, where [T] is a top-level theorem or a
constructor of an inductively defined proposition (i.e., anything
whose type is an implication). As a shorthand, we can write
Hint Constructors c.
to tell Coq to do a [Hint Resolve] for _all_ of the constructors
from the inductive definition of [c].
It is also sometimes necessary to add
Hint Unfold d.
where [d] is a defined symbol, so that [auto] knows to expand
uses of [d] and enable further possibilities for applying
lemmas that it knows about. *)
(** Here are some [Hint]s we will find useful. *)
Hint Constructors multi.
Hint Resolve beq_id_eq beq_id_false_not_eq.
(** Warning: Just as with Coq's other automation facilities, it is
easy to overuse [auto] and [eauto] and wind up with proofs that
are impossible to understand later!
Also, overuse of [eauto] can make proof scripts very slow. Get in
the habit of using [auto] most of the time and [eauto] only when
necessary.
For much more detailed information about using [auto] and [eauto],
see the chapter [UseAuto]. *)
(* ###################################################################### *)
(** ** The [Proof with] Tactic *)
(** If you start a proof by saying [Proof with (tactic)] instead of
just [Proof], then writing [...] instead of [.] after a tactic in
the body of the proof will try to solve all generated subgoals
with [tactic] (and fail if this doesn't work).
One common use of this facility is "[Proof with auto]" (or
[eauto]). We'll see many examples of this later in the file. *)
(* ###################################################################### *)
(** ** The [solve by inversion] Tactic *)
(** Here's another nice automation feature: it often arises that the
context contains a contradictory assumption and we want to use
[inversion] on it to solve the goal. We'd like to be able to say
to Coq, "find a contradictory assumption and invert it" without
giving its name explicitly.
Doing [solve by inversion] will find a hypothesis that can be
inverted to solve the goal, if there is one. The tactics [solve
by inversion 2] and [solve by inversion 3] are slightly fancier
versions which will perform two or three inversions in a row, if
necessary, to solve the goal.
(These tactics are not actually built into Coq -- their
definitions are in [Sflib].)
Caution: Overuse of [solve by inversion] can lead to slow proof
scripts. *)
(* ###################################################################### *)
(** ** The [try solve] Tactic *)
(** If [t] is a tactic, then [try solve [t]] is a tactic that
- if [t] solves the goal, behaves just like [t], or
- if [t] cannot completely solve the goal, does
nothing.
More generally, [try solve [t1 | t2 | ...]] will try to solve the
goal by using [t1], [t2], etc. If none of them succeeds in
completely solving the goal, then [try solve [t1 | t2 | ...]] does
nothing. *)
(* ###################################################################### *)
(** ** The [f_equal] Tactic *)
(** [f_equal] replaces a goal of the form [f x1 x2 ... xn = f y1 y2
... yn], where [f] is some function, with the subgoals [x1 = y1],
[x2 = y2],...,[xn = yn]. It is useful for avoiding explicit
rewriting steps, and often the generated subgoals can be quickly
cleared by [auto]. This tactic is not fundamental, in the sense
that it can always be replaced by a sequence of [assert]s.
However in some cases it can be very handy. *)
(* ###################################################################### *)
(** ** The [normalize] Tactic *)
(** When experimenting with definitions of programming languages in
Coq, we often want to see what a particular concrete term steps
to -- i.e., we want to find proofs for goals of the form [t ==>*
t'], where [t] is a completely concrete term and [t'] is unknown.
These proofs are simple but repetitive to do by hand. Consider for
example reducing an arithmetic expression using the small-step
relation [astep] defined in the previous chapter: *)
Definition amultistep st := multi (astep st).
Notation " t '/' st '==>a*' t' " := (amultistep st t t')
(at level 40, st at level 39).
Example astep_example1 :
(APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state
==>a* (ANum 15).
Proof.
apply multi_step with (APlus (ANum 3) (ANum 12)).
apply AS_Plus2.
apply av_num.
apply AS_Mult.
apply multi_step with (ANum 15).
apply AS_Plus.
apply multi_refl.
Qed.
(** We repeatedly applied [multi_step] until we got to a normal
form. The proofs that the intermediate steps are possible are
simple enough that [auto], with appropriate hints, can solve
them. *)
Hint Constructors astep aval.
Example astep_example1' :
(APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state
==>a* (ANum 15).
Proof.
eapply multi_step. auto. simpl.
eapply multi_step. auto. simpl.
apply multi_refl.
Qed.
(** The following custom [Tactic Notation] definition captures this
pattern. In addition, before each [multi_step] we print out the
current goal, so that the user can follow how the term is being
evaluated. *)
Tactic Notation "print_goal" := match goal with |- ?x => idtac x end.
Tactic Notation "normalize" :=
repeat (print_goal; eapply multi_step ;
[ (eauto 10; fail) | (instantiate; simpl)]);
apply multi_refl.
Example astep_example1'' :
(APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state
==>a* (ANum 15).
Proof.
normalize.
(* At this point in the proof script, the Coq response shows
a trace of how the expression evaluated.
(APlus (ANum 3) (AMult (ANum 3) (ANum 4)) / empty_state ==>a* ANum 15)
(multi (astep empty_state) (APlus (ANum 3) (ANum 12)) (ANum 15))
(multi (astep empty_state) (ANum 15) (ANum 15))
*)
Qed.
(** The [normalize] tactic also provides a simple way to calculate
what the normal form of a term is, by proving a goal with an
existential variable in it. *)
Example astep_example1''' : exists e',
(APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state
==>a* e'.
Proof.
eapply ex_intro. normalize.
(* This time the trace will be:
(APlus (ANum 3) (AMult (ANum 3) (ANum 4)) / empty_state ==>a* ??)
(multi (astep empty_state) (APlus (ANum 3) (ANum 12)) ??)
(multi (astep empty_state) (ANum 15) ??)
where ?? is the variable ``guessed'' by eapply.
*)
Qed.
(** **** Exercise: 1 star (normalize_ex) *)
(* EXPECTED *)
Theorem normalize_ex : exists e',
(AMult (ANum 3) (AMult (ANum 2) (ANum 1))) / empty_state
==>a* e'.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star, optional (normalize_ex') *)
(* EXPECTED *)
(** This time prove it by using [apply] instead of [eapply]. *)
Theorem normalize_ex' : exists e',
(AMult (ANum 3) (AMult (ANum 2) (ANum 1))) / empty_state
==>a* e'.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ###################################################################### *)
(** * Typed Arithmetic Expressions *)
(** To motivate the discussion of type systems, let's begin as
usual with an extremely simple toy language. We want it to have
the potential for programs "going wrong" because of runtime type
errors, so we need something a tiny bit more complex than the
language of constants and addition that we used in chapter
[Smallstep]: a single kind of data (just numbers) is too simple,
but just two kinds (numbers and booleans) already gives us enough
material to tell an interesting story.
The language definition is completely routine. The only thing to
notice is that we are _not_ using the [asnum]/[aslist] trick that
we used in chapter [ImpList] to make all the operations total by
forcibly coercing the arguments to [+] (for example) into numbers.
Instead, we simply let terms get stuck if they try to use an
operator with the wrong kind of operands: the [step] relation
doesn't relate them to anything. *)
(* ###################################################################### *)
(** ** Syntax *)
(** Informally:
t ::= true
| false
| if t then t else t
| 0
| succ t
| pred t
| iszero t
Formally:
*)
Inductive tm : Type :=
| ttrue : tm
| tfalse : tm
| tif : tm -> tm -> tm -> tm
| tzero : tm
| tsucc : tm -> tm
| tpred : tm -> tm
| tiszero : tm -> tm.
(** _Values_ are [true], [false], and numeric values... *)
Inductive bvalue : tm -> Prop :=
| bv_true : bvalue ttrue
| bv_false : bvalue tfalse.
Inductive nvalue : tm -> Prop :=
| nv_zero : nvalue tzero
| nv_succ : forall t, nvalue t -> nvalue (tsucc t).
Definition value (t:tm) := bvalue t \/ nvalue t.
Hint Constructors bvalue nvalue.
Hint Unfold value.
(* ###################################################################### *)
(** ** Operational Semantics *)
(** Informally:
------------------------------ (ST_IfTrue)
if true then t1 else t2 ==> t1
------------------------------- (ST_IfFalse)
if false then t1 else t2 ==> t2
t1 ==> t1'
------------------------- (ST_If)
if t1 then t2 else t3 ==>
if t1' then t2 else t3
t1 ==> t1'
-------------------- (ST_Succ)
succ t1 ==> succ t1'
------------ (ST_PredZero)
pred 0 ==> 0
numeric value v1
--------------------- (ST_PredSucc)
pred (succ v1) ==> v1
t1 ==> t1'
-------------------- (ST_Pred)
pred t1 ==> pred t1'
----------------- (ST_IszeroZero)
iszero 0 ==> true
numeric value v1
-------------------------- (ST_IszeroSucc)
iszero (succ v1) ==> false
t1 ==> t1'
------------------------ (ST_Iszero)
iszero t1 ==> iszero t1'
Formally:
*)
Reserved Notation "t1 '==>' t2" (at level 40).
Inductive step : tm -> tm -> Prop :=
| ST_IfTrue : forall t1 t2,
(tif ttrue t1 t2) ==> t1
| ST_IfFalse : forall t1 t2,
(tif tfalse t1 t2) ==> t2
| ST_If : forall t1 t1' t2 t3,
t1 ==> t1' ->
(tif t1 t2 t3) ==> (tif t1' t2 t3)
| ST_Succ : forall t1 t1',
t1 ==> t1' ->
(tsucc t1) ==> (tsucc t1')
| ST_PredZero :
(tpred tzero) ==> tzero
| ST_PredSucc : forall t1,
nvalue t1 ->
(tpred (tsucc t1)) ==> t1
| ST_Pred : forall t1 t1',
t1 ==> t1' ->
(tpred t1) ==> (tpred t1')
| ST_IszeroZero :
(tiszero tzero) ==> ttrue
| ST_IszeroSucc : forall t1,
nvalue t1 ->
(tiszero (tsucc t1)) ==> tfalse
| ST_Iszero : forall t1 t1',
t1 ==> t1' ->
(tiszero t1) ==> (tiszero t1')
where "t1 '==>' t2" := (step t1 t2).
Tactic Notation "step_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "ST_IfTrue" | Case_aux c "ST_IfFalse" | Case_aux c "ST_If"
| Case_aux c "ST_Succ" | Case_aux c "ST_PredZero"
| Case_aux c "ST_PredSucc" | Case_aux c "ST_Pred"
| Case_aux c "ST_IszeroZero" | Case_aux c "ST_IszeroSucc"
| Case_aux c "ST_Iszero" ].
Hint Constructors step.
(** Notice that the [step] relation doesn't care about whether
expressions make global sense -- it just checks that the operation
in the _next_ reduction step is being applied to the right kinds
of operands. For example, the term [succ true] (i.e., [tsucc
ttrue] in the formal syntax) cannot take a step, but the
almost-as-obviously-nonsensical term
succ (if true then true else true)
can take _one_ step. *)
(* ###################################################################### *)
(** ** Normal Forms and Values *)
(** The first interesting thing about the [step] relation in this
language is that the strong progress theorem from the Smallstep
chapter fails! That is, there are terms that are normal
forms (they can't take a step) but not values (because we have not
included them in our definition of possible "results of
evaluation"). Such terms are _stuck_. *)
Notation step_normal_form := (normal_form step).
Definition stuck (t:tm) : Prop :=
step_normal_form t /\ ~ value t.
Hint Unfold stuck.
(** **** Exercise: 2 stars (some_term_is_stuck) *)
(* EXPECTED *)
Example some_term_is_stuck :
exists t, stuck t.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** However, although values and normal forms are not the same in this
language, the former set is included in the latter. This is
important because it shows we did not accidentally define things
so that some value could still take a step. *)
(** **** Exercise: 3 stars, optional (value_is_nf) *)
(* EXPECTED *)
(** Hint: You will reach a point in this proof where you need to
use an induction to reason about a term that is known to be a
numeric value. This induction can be performed either over the
term itself or over the evidence that it is a numeric value. The
proof goes through in either case, but you will find that one way
is quite a bit shorter than the other. For the sake of the
exercise, try to complete the proof both ways. *)
Lemma value_is_nf : forall t,
value t -> step_normal_form t.
Proof.
Lemma bvalue_is_nf : step_normal_form (ttrue).
Proof. unfold normal_form, not; inversion 1;
match reverse goal with
[H: step _ _ |- _ ] =>
inversion H; subst
end.
Qed.
(* Lemma nvalue_is_nf : step_normal_form ( *)
(* induction 1; unfold normal_form. *)
Admitted.
(** [] *)
(** **** Exercise: 3 stars, optional (step_deterministic) *)
(** Using [value_is_nf], we can show that the [step] relation is
also deterministic... *)
Theorem step_deterministic:
deterministic step.
Proof with eauto.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ###################################################################### *)
(** ** Typing *)
(** The next critical observation about this language is that,
although there are stuck terms, they are all "nonsensical", mixing
booleans and numbers in a way that we don't even _want_ to have a
meaning. We can easily exclude such ill-typed terms by defining a
_typing relation_ that relates terms to the types (either numeric
or boolean) of their final results. *)
Inductive ty : Type :=
| TBool : ty
| TNat : ty.
(** In informal notation, the typing relation is often written
[|- t : T], pronounced "[t] has type [T]." The [|-] symbol is
called a "turnstile". (Below, we're going to see richer typing
relations where an additional "context" argument is written to the
left of the turnstile. Here, the context is always empty.) *)
(**
-------------- (T_True)
|- true : Bool
--------------- (T_False)
|- false : Bool
|- t1 : Bool |- t2 : T |- t3 : T
-------------------------------------- (T_If)
|- if t1 then t2 else t3 : T
---------- (T_Zero)
|- 0 : Nat
|- t1 : Nat
---------------- (T_Succ)
|- succ t1 : Nat
|- t1 : Nat
---------------- (T_Pred)
|- pred t1 : Nat
|- t1 : Nat
------------------- (T_IsZero)
|- iszero t1 : Bool
*)
Inductive has_type : tm -> ty -> Prop :=
| T_True :
has_type ttrue TBool
| T_False :
has_type tfalse TBool
| T_If : forall t1 t2 t3 T,
has_type t1 TBool ->
has_type t2 T ->
has_type t3 T ->
has_type (tif t1 t2 t3) T
| T_Zero :
has_type tzero TNat
| T_Succ : forall t1,
has_type t1 TNat ->
has_type (tsucc t1) TNat
| T_Pred : forall t1,
has_type t1 TNat ->
has_type (tpred t1) TNat
| T_Iszero : forall t1,
has_type t1 TNat ->
has_type (tiszero t1) TBool.
Tactic Notation "has_type_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "T_True" | Case_aux c "T_False" | Case_aux c "T_If"
| Case_aux c "T_Zero" | Case_aux c "T_Succ" | Case_aux c "T_Pred"
| Case_aux c "T_Iszero" ].
Hint Constructors has_type.
(* ###################################################################### *)
(** *** Examples *)
(** It's important to realize that the typing relation is a
_conservative_ (or _static_) approximation: it does not calculate
the type of the normal form of a term. *)
Example has_type_1 :
has_type (tif tfalse tzero (tsucc tzero)) TNat.
Proof.
apply T_If.
apply T_False.
apply T_Zero.
apply T_Succ.
apply T_Zero.
Qed.
(** (Since we've included all the constructors of the typing relation
in the hint database, the [auto] tactic can actually find this
proof automatically.) *)
Example has_type_not :
~ has_type (tif tfalse tzero ttrue) TBool.
Proof.
intros Contra. solve by inversion 2. Qed.
(** **** Exercise: 1 star (succ_hastype_nat__hastype_nat) *)
(* EXPECTED *)
Example succ_hastype_nat__hastype_nat : forall t,
has_type (tsucc t) TNat ->
has_type t TNat.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ###################################################################### *)
(** ** Progress *)
(** The typing relation enjoys two critical properties. The first is
that well-typed normal forms are values (i.e., not stuck). *)
(** **** Exercise: 3 stars, recommended (finish_progress_informal) *)
(** Complete the following proof: *)
(** _Theorem_: If [|- t : T], then either [t] is a value or else
[t ==> t'] for some [t']. *)
(** _Proof_: By induction on a derivation of [|- t : T].
- If the last rule in the derivation is [T_If], then [t = if t1
then t2 else t3], with [|- t1 : Bool], [|- t2 : T] and [|- t3
: T]. By the IH, either [t1] is a value or else [t1] can step
to some [t1'].
- If [t1] is a value, then it is either an [nvalue] or a
[bvalue]. But it cannot be an [nvalue], because we know
[|- t1 : Bool] and there are no rules assigning type
[Bool] to any term that could be an [nvalue]. So [t1]
is a [bvalue] -- i.e., it is either [true] or [false].
If [t1 = true], then [t] steps to [t2] by [ST_IfTrue],
while if [t1 = false], then [t] steps to [t3] by
[ST_IfFalse]. Either way, [t] can step, which is what
we wanted to show.
- If [t1] itself can take a step, then, by [ST_If], so can
[t].
(* FILL IN HERE *)
[]
*)
(** **** Exercise: 3 stars (finish_progress) *)
(* EXPECTED *)
Theorem progress : forall t T,
has_type t T ->
value t \/ exists t', t ==> t'.
Proof with auto.
intros t T HT.
has_type_cases (induction HT) Case...
(* The cases that were obviously values, like T_True and
T_False, were eliminated immediately by auto *)
Case "T_If".
right. inversion IHHT1; clear IHHT1.
SCase "t1 is a value". inversion H; clear H.
SSCase "t1 is a bvalue". inversion H0; clear H0.
SSSCase "t1 is ttrue".
exists t2...
SSSCase "t1 is tfalse".
exists t3...
SSCase "t1 is an nvalue".
solve by inversion 2. (* on H and HT1 *)
SCase "t1 can take a step".
inversion H as [t1' H1].
exists (tif t1' t2 t3)...
(* FILL IN HERE *) Admitted.
(** [] *)
(** This is more interesting than the strong progress theorem that we
saw in the Smallstep chapter, where _all_ normal forms were
values. Here, a term can be stuck, but only if it is ill
typed. *)
(** **** Exercise: 1 star (step_review) *)
(* EXPECTED *)
(** Quick review. Answer _true_ or _false_. In this language...
- Every well-typed normal form is a value.
- Every value is a normal form.
- The single-step evaluation relation is
a partial function (i.e., it is deterministic).
- The single-step evaluation relation is a _total_ function.
*)
(** [] *)
(* ###################################################################### *)
(** ** Type Preservation *)
(** The second critical property of typing is that, when a well-typed
term takes a step, the result is also a well-typed term.
This theorem is often called the _subject reduction_ property,
because it tells us what happens when the "subject" of the typing
relation is reduced. This terminology comes from thinking of
typing statements as sentences, where the term is the subject and
the type is the predicate. *)
(** **** Exercise: 3 stars, recommended (finish_preservation_informal) *)
(** Complete the following proof: *)
(** _Theorem_: If [|- t : T] and [t ==> t'], then [|- t' : T]. *)
(** _Proof_: By induction on a derivation of [|- t : T].
- If the last rule in the derivation is [T_If], then [t = if t1
then t2 else t3], with [|- t1 : Bool], [|- t2 : T] and [|- t3
: T].
Inspecting the rules for the small-step reduction relation and
remembering that [t] has the form [if ...], we see that the
only ones that could have been used to prove [t ==> t'] are
[ST_IfTrue], [ST_IfFalse], or [ST_If].
- If the last rule was [ST_IfTrue], then [t' = t2]. But we
know that [|- t2 : T], so we are done.
- If the last rule was [ST_IfFalse], then [t' = t3]. But we
know that [|- t3 : T], so we are done.
- If the last rule was [ST_If], then [t' = if t1' then t2
else t3], where [t1 ==> t1']. We know [|- t1 : Bool] so,
by the IH, [|- t1' : Bool]. The [T_If] rule then gives us
[|- if t1' then t2 else t3 : T], as required.
(* FILL IN HERE *)
[]
*)
(** **** Exercise: 2 stars (finish_preservation) *)
Theorem preservation : forall t t' T,
has_type t T ->
t ==> t' ->
has_type t' T.
Proof with auto.
intros t t' T HT HE.
generalize dependent t'.
has_type_cases (induction HT) Case;
(* every case needs to introduce a couple of things *)
intros t' HE;
(* and we can deal with several impossible
cases all at once *)
try (solve by inversion).
Case "T_If". inversion HE; subst.
SCase "ST_IFTrue". assumption.
SCase "ST_IfFalse". assumption.
SCase "ST_If". apply T_If; try assumption.
apply IHHT1; assumption.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars (preservation_alternate_proof) *)
(* EXPECTED *)
(** Now prove the same property again by induction on the
_evaluation_ derivation instead of on the typing derivation.
Begin by carefully reading and thinking about the first few
lines of the above proof to make sure you understand what
each one is doing. The set-up for this proof is similar, but
not exactly the same. *)
Theorem preservation' : forall t t' T,
has_type t T ->
t ==> t' ->
has_type t' T.
Proof with eauto.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ###################################################################### *)
(** ** Type Soundness *)
(** Putting progress and preservation together, we can see that a
well-typed term can _never_ reach a stuck state. *)
Definition multistep := (multi step).
Notation "t1 '==>*' t2" := (multistep t1 t2) (at level 40).
Corollary soundness : forall t t' T,
has_type t T ->
t ==>* t' ->
~(stuck t').
Proof.
intros t t' T HT P. induction P; intros [R S].
destruct (progress x T HT); auto.
apply IHP. apply (preservation x y T HT H).
unfold stuck. split; auto. Qed.
(* ###################################################################### *)
(** ** Additional Exercises *)
(** **** Exercise: 2 stars, recommended (subject_expansion) *)
(* EXPECTED *)
(** Having seen the subject reduction property, it is reasonable to
wonder whether the opposity property -- subject _expansion_ --
also holds. That is, is it always the case that, if [t ==> t']
and [has_type t' T], then [has_type t T]? If so, prove it. If
not, give a counter-example. (You do not need to prove your
counter-example in Coq, but feel free to do so if you like.)
(* FILL IN HERE *)
[]
*)
(** **** Exercise: 2 stars (variation1) *)
(* EXPECTED *)
(** Suppose we add the following two new rules to the reduction
relation:
| ST_PredTrue :
(tpred ttrue) ==> (tpred tfalse)
| ST_PredFalse :
(tpred tfalse) ==> (tpred ttrue)
Which of the following properties remain true in the presence
of these rules? 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 (variation2) *)
(* EXPECTED *)
(** Suppose, instead, that we add this new rule to the typing relation:
| T_IfFunny : forall t2 t3,
has_type t2 TNat ->
has_type (tif ttrue t2 t3) TNat
Which of the following properties remain true in the presence of
this rule? (Answer in the same style as above.)
- Determinism of [step]
- Progress
- Preservation
[]
*)
(** **** Exercise: 2 stars (variation3) *)
(** Suppose, instead, that we add this new rule to the typing relation:
| T_SuccBool : forall t,
has_type t TBool ->
has_type (tsucc t) TBool
Which of the following properties remain true in the presence of
this rule? (Answer in the same style as above.)
- Determinism of [step]
- Progress
- Preservation
[]
*)
(** **** Exercise: 2 stars (variation4) *)
(** Suppose, instead, that we add this new rule to the [step] relation:
| ST_Funny1 : forall t2 t3,
(tif ttrue t2 t3) ==> t3
Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
[]
*)
(** **** Exercise: 2 stars (variation5) *)
(** Suppose instead that we add this rule:
| ST_Funny2 : forall t1 t2 t2' t3,
t2 ==> t2' ->
(tif t1 t2 t3) ==> (tif t1 t2' t3)
Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
[]
*)
(** **** Exercise: 2 stars (variation6) *)
(** Suppose instead that we add this rule:
| ST_Funny3 :
(tpred tfalse) ==> (tpred (tpred tfalse))
Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
[]
*)
(** **** Exercise: 2 stars (variation7) *)
(** Suppose instead that we add this rule:
| T_Funny4 :
has_type tzero TBool
]]
Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
[]
*)
(** **** Exercise: 2 stars (variation8) *)
(** Suppose instead that we add this rule:
| T_Funny5 :
has_type (tpred tzero) TBool
]]
Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
[]
*)
(** **** Exercise: 3 stars, optional (more_variations) *)
(** Make up some exercises of your own along the same lines as
the ones above. Try to find ways of selectively breaking
properties -- i.e., ways of changing the definitions that
break just one of the properties and leave the others alone.
[]
*)
(** **** Exercise: 1 star (remove_predzero) *)
(** The evaluation rule [E_PredZero] is a bit counter-intuitive: we
might feel that it makes more sense for the predecessor of zero to
be undefined, rather than being defined to be zero. Can we
achieve this simply by removing the rule from the definition of
[step]? Would doing so create any problems elsewhere?
(* FILL IN HERE *)
[] *)
(** **** Exercise: 4 stars, optional (prog_pres_bigstep) *)
(** Suppose our evaluation relation is defined in the big-step style.
What are the appropriate analogs of the progress and preservation
properties?
(* FILL IN HERE *)
[]
*)