TypesType Systems
Require Import Coq.Arith.Arith.
Require Import SfLib.
Require Import Maps.
Require Import Imp.
Require Import Smallstep.
(* INSTRUCTORS. : APT: best place I could find for these without
requiring new Auto *)
Hint Constructors multi.
Our next major topic 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).
Typed Arithmetic Expressions
Syntax
t ::= true
| false
| if t then t else t
| 0
| succ t
| pred t
| iszero t
Formally:
| false
| if t then t else t
| 0
| succ t
| pred t
| iszero t
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 : ∀t, nvalue t → nvalue (tsucc t).
Definition value (t:tm) := bvalue t ∨ nvalue t.
Operational Semantics
(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' |
Reserved Notation "t1 '⇒' t2" (at level 40).
Inductive step : tm → tm → Prop :=
| ST_IfTrue : ∀t1 t2,
(tif ttrue t1 t2) ⇒ t1
| ST_IfFalse : ∀t1 t2,
(tif tfalse t1 t2) ⇒ t2
| ST_If : ∀t1 t1' t2 t3,
t1 ⇒ t1' →
(tif t1 t2 t3) ⇒ (tif t1' t2 t3)
| ST_Succ : ∀t1 t1',
t1 ⇒ t1' →
(tsucc t1) ⇒ (tsucc t1')
| ST_PredZero :
(tpred tzero) ⇒ tzero
| ST_PredSucc : ∀t1,
nvalue t1 →
(tpred (tsucc t1)) ⇒ t1
| ST_Pred : ∀t1 t1',
t1 ⇒ t1' →
(tpred t1) ⇒ (tpred t1')
| ST_IszeroZero :
(tiszero tzero) ⇒ ttrue
| ST_IszeroSucc : ∀t1,
nvalue t1 →
(tiszero (tsucc t1)) ⇒ tfalse
| ST_Iszero : ∀t1 t1',
t1 ⇒ t1' →
(tiszero t1) ⇒ (tiszero t1')
where "t1 '⇒' t2" := (step t1 t2).
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 a step (once, before becoming stuck).
Normal Forms and Values
Notation step_normal_form := (normal_form step).
Definition stuck (t:tm) : Prop :=
step_normal_form t ∧ ¬ value t.
Hint Unfold stuck.
☐
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, advanced (value_is_nf)
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.
☐
Exercise: 3 stars, optional (step_deterministic)
Using value_is_nf, we can show that the step relation is also deterministic...
☐
Typing
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 |
Reserved Notation "'⊢' t '∈' T" (at level 40).
Inductive has_type : tm → ty → Prop :=
| T_True :
⊢ ttrue ∈ TBool
| T_False :
⊢ tfalse ∈ TBool
| T_If : ∀t1 t2 t3 T,
⊢ t1 ∈ TBool →
⊢ t2 ∈ T →
⊢ t3 ∈ T →
⊢ tif t1 t2 t3 ∈ T
| T_Zero :
⊢ tzero ∈ TNat
| T_Succ : ∀t1,
⊢ t1 ∈ TNat →
⊢ tsucc t1 ∈ TNat
| T_Pred : ∀t1,
⊢ t1 ∈ TNat →
⊢ tpred t1 ∈ TNat
| T_Iszero : ∀t1,
⊢ t1 ∈ TNat →
⊢ tiszero t1 ∈ TBool
where "'⊢' t '∈' T" := (has_type t T).
Examples
(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 :
¬ (⊢ tif tfalse tzero ttrue ∈ TBool).
Proof.
intros Contra. solve by inversion 2. Qed.
intros Contra. solve by inversion 2. Qed.
Example succ_hastype_nat__hastype_nat : ∀t,
⊢ tsucc t ∈ TNat →
⊢ t ∈ TNat.
Proof.
(* FILL IN HERE *) Admitted.
⊢ tsucc t ∈ TNat →
⊢ t ∈ TNat.
Proof.
(* FILL IN HERE *) Admitted.
☐
Canonical forms
Lemma bool_canonical : ∀t,
⊢ t ∈ TBool → value t → bvalue t.
Proof.
intros t HT HV.
inversion HV; auto.
induction H; inversion HT; auto.
Qed.
intros t HT HV.
inversion HV; auto.
induction H; inversion HT; auto.
Qed.
Lemma nat_canonical : ∀t,
⊢ t ∈ TNat → value t → nvalue t.
Proof.
intros t HT HV.
inversion HV.
inversion H; subst; inversion HT.
auto.
Qed.
intros t HT HV.
inversion HV.
inversion H; subst; inversion HT.
auto.
Qed.
Progress
Exercise: 3 stars (finish_progress)
Complete the formal proof of the progress property. (Make sure you understand the informal proof fragment in the following exercise before starting — this will save you a lot of time.)
Proof with auto.
intros t T HT.
induction HT...
(* The cases that were obviously values, like T_True and
T_False, were eliminated immediately by auto *)
- (* T_If *)
right. inversion IHHT1; clear IHHT1.
+ (* t1 is a value *)
apply (bool_canonical t1 HT1) in H.
inversion H; subst; clear H.
∃t2...
∃t3...
+ (* t1 can take a step *)
inversion H as [t1' H1].
∃(tif t1' t2 t3)...
(* FILL IN HERE *) Admitted.
intros t T HT.
induction HT...
(* The cases that were obviously values, like T_True and
T_False, were eliminated immediately by auto *)
- (* T_If *)
right. inversion IHHT1; clear IHHT1.
+ (* t1 is a value *)
apply (bool_canonical t1 HT1) in H.
inversion H; subst; clear H.
∃t2...
∃t3...
+ (* t1 can take a step *)
inversion H as [t1' H1].
∃(tif t1' t2 t3)...
(* FILL IN HERE *) Admitted.
☐
Exercise: 3 stars, advanced (finish_progress_informal)
Complete the corresponding informal proof:- 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 by the canonical forms lemmas
and the fact that ⊢ t1 ∈ Bool we have that 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.
- If t1 is a value, then by the canonical forms lemmas
and the fact that ⊢ t1 ∈ Bool we have that 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.
☐
Exercise: 1 star (step_review)
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
Exercise: 2 stars (finish_preservation)
Complete the formal proof of the preservation property. (Again, make sure you understand the informal proof fragment in the following exercise first.)
Proof with auto.
intros t t' T HT HE.
generalize dependent t'.
induction HT;
(* 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).
- (* T_If *) inversion HE; subst; clear HE.
+ (* ST_IFTrue *) assumption.
+ (* ST_IfFalse *) assumption.
+ (* ST_If *) apply T_If; try assumption.
apply IHHT1; assumption.
(* FILL IN HERE *) Admitted.
intros t t' T HT HE.
generalize dependent t'.
induction HT;
(* 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).
- (* T_If *) inversion HE; subst; clear HE.
+ (* ST_IFTrue *) assumption.
+ (* ST_IfFalse *) assumption.
+ (* ST_If *) apply T_If; try assumption.
apply IHHT1; assumption.
(* FILL IN HERE *) Admitted.
☐
Theorem: If ⊢ t ∈ T and t ⇒ t', then ⊢ t' ∈ T.
Proof: By induction on a derivation of ⊢ t ∈ T.
(* FILL IN HERE *)
☐
Exercise: 3 stars, advanced (finish_preservation_informal)
Complete the following proof:- 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.
- 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.
- If the last rule was ST_IfTrue, then t' = t2. But we
know that ⊢ t2 ∈ T, so we are done.
☐
Exercise: 3 stars (preservation_alternate_proof)
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' : ∀t t' T,
⊢ t ∈ T →
t ⇒ t' →
⊢ t' ∈ T.
Proof with eauto.
(* FILL IN HERE *) Admitted.
☐
Type Soundness
Definition multistep := (multi step).
Notation "t1 '⇒*' t2" := (multistep t1 t2) (at level 40).
Corollary soundness : ∀t t' T,
⊢ 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.
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.
Aside: the normalize Tactic
Definition amultistep st := multi (astep st).
Notation " t '/' st '⇒a×' t' " := (amultistep st t t')
(at level 40, st at level 39).
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 apply multi_step until we get 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''' : ∃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.
Theorem normalize_ex : ∃e',
(AMult (ANum 3) (AMult (ANum 2) (ANum 1))) / empty_state
⇒a× e'.
Proof.
(* FILL IN HERE *) Admitted.
(AMult (ANum 3) (AMult (ANum 2) (ANum 1))) / empty_state
⇒a× e'.
Proof.
(* FILL IN HERE *) Admitted.
☐
Exercise: 1 star, optional (normalize_ex')
For comparison, prove it using apply instead of eapply.Theorem normalize_ex' : ∃e',
(AMult (ANum 3) (AMult (ANum 2) (ANum 1))) / empty_state
⇒a× e'.
Proof.
(* FILL IN HERE *) Admitted.
☐
Additional Exercises
Exercise: 2 stars, recommended (subject_expansion)
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 ⊢ t' ∈ T, then ⊢ 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.)☐
Exercise: 2 stars (variation1)
Suppose, that we add this new rule to the typing relation:
| T_SuccBool : ∀t,
⊢ t ∈ TBool →
⊢ tsucc t ∈ TBool
Which of the following properties 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.
⊢ t ∈ TBool →
⊢ tsucc t ∈ TBool
- Determinism of step
- Progress
- Preservation
Exercise: 2 stars (variation2)
Suppose, instead, that we add this new rule to the step relation:
| ST_Funny1 : ∀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.
(tif ttrue t2 t3) ⇒ t3
Exercise: 2 stars, optional (variation3)
Suppose instead that we add this rule:
| ST_Funny2 : ∀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.
t2 ⇒ t2' →
(tif t1 t2 t3) ⇒ (tif t1 t2' t3)
Exercise: 2 stars, optional (variation4)
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.
(tpred tfalse) ⇒ (tpred (tpred tfalse))
Exercise: 2 stars, optional (variation5)
Suppose instead that we add this rule:
| T_Funny4 :
⊢ tzero ∈ TBool
Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
⊢ tzero ∈ TBool
Exercise: 2 stars, optional (variation6)
Suppose instead that we add this rule:
| T_Funny5 :
⊢ 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.
⊢ tpred tzero ∈ TBool
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?☐
Exercise: 4 stars, advanced (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?☐