TypesType Systems
Require Export Smallstep.
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.
Hint Constructors bvalue nvalue.
Hint Unfold value.
Hint Unfold extend.
Hint Unfold value.
Hint Unfold extend.
Operational Semantics
(ST_IfTrue) | |
if true then t_{1} else t_{2} ⇒ t_{1} |
(ST_IfFalse) | |
if false then t_{1} else t_{2} ⇒ t_{2} |
t_{1} ⇒ t_{1}' | (ST_If) |
if t_{1} then t_{2} else t_{3} ⇒ | |
if t_{1}' then t_{2} else t_{3} |
t_{1} ⇒ t_{1}' | (ST_Succ) |
succ t_{1} ⇒ succ t_{1}' |
(ST_PredZero) | |
pred 0 ⇒ 0 |
numeric value v_{1} | (ST_PredSucc) |
pred (succ v_{1}) ⇒ v_{1} |
t_{1} ⇒ t_{1}' | (ST_Pred) |
pred t_{1} ⇒ pred t_{1}' |
(ST_IszeroZero) | |
iszero 0 ⇒ true |
numeric value v_{1} | (ST_IszeroSucc) |
iszero (succ v_{1}) ⇒ false |
t_{1} ⇒ t_{1}' | (ST_Iszero) |
iszero t_{1} ⇒ iszero t_{1}' |
Reserved Notation "t_{1} '⇒' t_{2}" (at level 40).
Inductive step : tm → tm → Prop :=
| ST_IfTrue : ∀t_{1} t_{2},
(tif ttrue t_{1} t_{2}) ⇒ t_{1}
| ST_IfFalse : ∀t_{1} t_{2},
(tif tfalse t_{1} t_{2}) ⇒ t_{2}
| ST_If : ∀t_{1} t_{1}' t_{2} t_{3},
t_{1} ⇒ t_{1}' →
(tif t_{1} t_{2} t_{3}) ⇒ (tif t_{1}' t_{2} t_{3})
| ST_Succ : ∀t_{1} t_{1}',
t_{1} ⇒ t_{1}' →
(tsucc t_{1}) ⇒ (tsucc t_{1}')
| ST_PredZero :
(tpred tzero) ⇒ tzero
| ST_PredSucc : ∀t_{1},
nvalue t_{1} →
(tpred (tsucc t_{1})) ⇒ t_{1}
| ST_Pred : ∀t_{1} t_{1}',
t_{1} ⇒ t_{1}' →
(tpred t_{1}) ⇒ (tpred t_{1}')
| ST_IszeroZero :
(tiszero tzero) ⇒ ttrue
| ST_IszeroSucc : ∀t_{1},
nvalue t_{1} →
(tiszero (tsucc t_{1})) ⇒ tfalse
| ST_Iszero : ∀t_{1} t_{1}',
t_{1} ⇒ t_{1}' →
(tiszero t_{1}) ⇒ (tiszero t_{1}')
where "t_{1} '⇒' t_{2}" := (step t_{1} t_{2}).
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.
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 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.
Example some_term_is_stuck :
∃t, stuck t.
∃t, stuck t.
Proof.
(* FILL IN HERE *) Admitted.
(* 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, 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.Lemma value_is_nf : ∀t,
value t → step_normal_form t.
Proof.
(* FILL IN HERE *) Admitted.
(* FILL IN HERE *) 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
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 |
⊢ t_{1} ∈ Bool ⊢ t_{2} ∈ T ⊢ t_{3} ∈ T | (T_If) |
⊢ if t_{1} then t_{2} else t_{3} ∈ T |
(T_Zero) | |
⊢ 0 ∈ Nat |
⊢ t_{1} ∈ Nat | (T_Succ) |
⊢ succ t_{1} ∈ Nat |
⊢ t_{1} ∈ Nat | (T_Pred) |
⊢ pred t_{1} ∈ Nat |
⊢ t_{1} ∈ Nat | (T_IsZero) |
⊢ iszero t_{1} ∈ Bool |
Reserved Notation "'⊢' t '∈' T" (at level 40).
Inductive has_type : tm → ty → Prop :=
| T_True :
⊢ ttrue ∈ TBool
| T_False :
⊢ tfalse ∈ TBool
| T_If : ∀t_{1} t_{2} t_{3} T,
⊢ t_{1} ∈ TBool →
⊢ t_{2} ∈ T →
⊢ t_{3} ∈ T →
⊢ tif t_{1} t_{2} t_{3} ∈ T
| T_Zero :
⊢ tzero ∈ TNat
| T_Succ : ∀t_{1},
⊢ t_{1} ∈ TNat →
⊢ tsucc t_{1} ∈ TNat
| T_Pred : ∀t_{1},
⊢ t_{1} ∈ TNat →
⊢ tpred t_{1} ∈ TNat
| T_Iszero : ∀t_{1},
⊢ t_{1} ∈ TNat →
⊢ tiszero t_{1} ∈ TBool
where "'⊢' t '∈' T" := (has_type t T).
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.
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
Example has_type_1 :
⊢ tif tfalse tzero (tsucc tzero) ∈ TNat.
Proof.
apply T_If.
apply T_False.
apply T_Zero.
apply T_Succ.
apply T_Zero.
Qed.
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 :
¬ (⊢ 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.
☐
Progress
Theorem progress : ∀t T,
⊢ t ∈ T →
value t ∨ ∃t', t ⇒ t'.
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.
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 "t_{1} is a value". inversion H; clear H.
SSCase "t_{1} is a bvalue". inversion H0; clear H0.
SSSCase "t_{1} is ttrue".
∃t_{2}...
SSSCase "t_{1} is tfalse".
∃t_{3}...
SSCase "t_{1} is an nvalue".
solve by inversion 2. (* on H and HT1 *)
SCase "t_{1} can take a step".
inversion H as [t_{1}' H1].
∃(tif t_{1}' t_{2} t_{3})...
(* FILL IN HERE *) Admitted.
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 "t_{1} is a value". inversion H; clear H.
SSCase "t_{1} is a bvalue". inversion H0; clear H0.
SSSCase "t_{1} is ttrue".
∃t_{2}...
SSSCase "t_{1} is tfalse".
∃t_{3}...
SSCase "t_{1} is an nvalue".
solve by inversion 2. (* on H and HT1 *)
SCase "t_{1} can take a step".
inversion H as [t_{1}' H1].
∃(tif t_{1}' t_{2} t_{3})...
(* 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 t_{1}
then t_{2} else t_{3}, with ⊢ t_{1} ∈ Bool, ⊢ t_{2} ∈ T and ⊢ t_{3}
∈ T. By the IH, either t_{1} is a value or else t_{1} can step
to some t_{1}'.
- If t_{1} is a value, then it is either an nvalue or a
bvalue. But it cannot be an nvalue, because we know
⊢ t_{1} ∈ Bool and there are no rules assigning type
Bool to any term that could be an nvalue. So t_{1}
is a bvalue — i.e., it is either true or false.
If t_{1} = true, then t steps to t_{2} by ST_IfTrue,
while if t_{1} = false, then t steps to t_{3} by
ST_IfFalse. Either way, t can step, which is what
we wanted to show.
- If t_{1} itself can take a step, then, by ST_If, so can t.
- If t_{1} is a value, then it is either an nvalue or a
bvalue. But it cannot be an nvalue, because we know
⊢ t_{1} ∈ Bool and there are no rules assigning type
Bool to any term that could be an nvalue. So t_{1}
is a bvalue — i.e., it is either true or false.
If t_{1} = true, then t steps to t_{2} by ST_IfTrue,
while if t_{1} = false, then t steps to t_{3} 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
Theorem preservation : ∀t t' T,
⊢ t ∈ T →
t ⇒ t' →
⊢ t' ∈ T.
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'.
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.
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.
☐
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 t_{1}
then t_{2} else t_{3}, with ⊢ t_{1} ∈ Bool, ⊢ t_{2} ∈ T and ⊢ t_{3}
∈ T.
- If the last rule was ST_IfTrue, then t' = t_{2}. But we
know that ⊢ t_{2} ∈ T, so we are done.
- If the last rule was ST_IfFalse, then t' = t_{3}. But we
know that ⊢ t_{3} ∈ T, so we are done.
- If the last rule was ST_If, then t' = if t_{1}' then t_{2} else t_{3}, where t_{1} ⇒ t_{1}'. We know ⊢ t_{1} ∈ Bool so, by the IH, ⊢ t_{1}' ∈ Bool. The T_If rule then gives us ⊢ if t_{1}' then t_{2} else t_{3} ∈ T, as required.
- If the last rule was ST_IfTrue, then t' = t_{2}. But we
know that ⊢ t_{2} ∈ 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.
☐
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.
☐
Type Soundness
Definition multistep := (multi step).
Notation "t_{1} '⇒*' t_{2}" := (multistep t_{1} t_{2}) (at level 40).
Notation "t_{1} '⇒*' t_{2}" := (multistep t_{1} t_{2}) (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.
Additional Exercises
Exercise: 2 stars (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 : ∀t_{2} t_{3},
(tif ttrue t_{2} t_{3}) ⇒ t_{3}
Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
(tif ttrue t_{2} t_{3}) ⇒ t_{3}
Exercise: 2 stars, optional (variation3)
Suppose instead that we add this rule:
| ST_Funny2 : ∀t_{1} t_{2} t_{2}' t_{3},
t_{2} ⇒ t_{2}' →
(tif t_{1} t_{2} t_{3}) ⇒ (tif t_{1} t_{2}' t_{3})
Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
t_{2} ⇒ t_{2}' →
(tif t_{1} t_{2} t_{3}) ⇒ (tif t_{1} t_{2}' t_{3})
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?☐
(* $Date: 2013-07-17 16:19:11 -0400 (Wed, 17 Jul 2013) $ *)