TypesType Systems

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 the simplest imaginable language, to introduce the basic ideas of types and typing rules and the fundamental theorems about type systems: type preservation and progress. In chapter Stlc we'll move on to the simply typed lambda-calculus, which lives at the core of every modern functional programming language (including Coq!).
Set Warnings "-notation-overridden,-parsing,-deprecated-hint-without-locality".
From Coq Require Import Arith.Arith.
From PLF Require Import Maps.
From PLF Require Import Smallstep.

Hint Constructors multi : core.

Typed Arithmetic Expressions

To motivate the discussion of type systems, let's begin as usual with a tiny toy language. We want it to have the potential for programs to go 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 (e.g., numbers) is too simple, but just two kinds (numbers and booleans) gives us enough material to tell an interesting story.
The language definition is completely routine.

Syntax

Here is the syntax, informally:
t ::= true
| false
| if t then t else t
| 0
| succ t
| pred t
| iszero t
And here it is formally:
Module TM.

Inductive tm : Type :=
| tru : tm
| fls : tm
| ite : tm tm tm tm
| zro : tm
| scc : tm tm
| prd : tm tm
| iszro : tm tm.

Declare Custom Entry tm.
Declare Scope tm_scope.
Notation "'true'" := true (at level 1): tm_scope.
Notation "'true'" := (tru) (in custom tm at level 0): tm_scope.
Notation "'false'" := false (at level 1): tm_scope.
Notation "'false'" := (fls) (in custom tm at level 0): tm_scope.
Notation "<{ e }>" := e (e custom tm at level 99): tm_scope.
Notation "( x )" := x (in custom tm, x at level 99): tm_scope.
Notation "x" := x (in custom tm at level 0, x constr at level 0): tm_scope.
Notation "'0'" := (zro) (in custom tm at level 0): tm_scope.
Notation "'0'" := 0 (at level 1): tm_scope.
Notation "'succ' x" := (scc x) (in custom tm at level 90, x custom tm at level 80): tm_scope.
Notation "'pred' x" := (prd x) (in custom tm at level 90, x custom tm at level 80): tm_scope.
Notation "'iszero' x" := (iszro x) (in custom tm at level 80, x custom tm at level 70): tm_scope.
Notation "'if' c 'then' t 'else' e" := (ite c t e)
(in custom tm at level 90, c custom tm at level 80,
t custom tm at level 80, e custom tm at level 80): tm_scope.
Local Open Scope tm_scope.
Values are <{true}>, <{false}>, and numeric values...
Inductive bvalue : tm Prop :=
| bv_True : bvalue <{ true }>
| bv_false : bvalue <{ false }>.

Inductive nvalue : tm Prop :=
| nv_0 : nvalue <{ 0 }>
| nv_succ : t, nvalue t nvalue <{ succ t }>.

Definition value (t : tm) := bvalue t nvalue t.

Hint Constructors bvalue nvalue : core.
Hint Unfold value : core.

Operational Semantics

Here is the single-step relation, first 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_Pred0) pred 0 --> 0
 numeric value v (ST_PredSucc) pred (succ v) --> v
 t1 --> t1' (ST_Pred) pred t1 --> pred t1'
 (ST_IsZero0) iszero 0 --> true
 numeric value v (ST_IszeroSucc) iszero (succ v) --> false
 t1 --> t1' (ST_Iszero) iszero t1 --> iszero t1'
... and then formally:
Reserved Notation "t '-->' t'" (at level 40).

Inductive step : tm tm Prop :=
| ST_IfTrue : t1 t2,
<{ if true then t1 else t2 }> --> t1
| ST_IfFalse : t1 t2,
<{ if false then t1 else t2 }> --> t2
| ST_If : c c' t2 t3,
c --> c'
<{ if c then t2 else t3 }> --> <{ if c' then t2 else t3 }>
| ST_Succ : t1 t1',
t1 --> t1'
<{ succ t1 }> --> <{ succ t1' }>
| ST_Pred0 :
<{ pred 0 }> --> <{ 0 }>
| ST_PredSucc : v,
nvalue v
<{ pred (succ v) }> --> v
| ST_Pred : t1 t1',
t1 --> t1'
<{ pred t1 }> --> <{ pred t1' }>
| ST_Iszero0 :
<{ iszero 0 }> --> <{ true }>
| ST_IszeroSucc : v,
nvalue v
<{ iszero (succ v) }> --> <{ false }>
| ST_Iszero : t1 t1',
t1 --> t1'
<{ iszero t1 }> --> <{ iszero t1' }>

where "t '-->' t'" := (step t t').

Hint Constructors step : core.
Notice that the step relation doesn't care about whether the expression being stepped makes 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 }> 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

The first interesting thing to notice about this step relation is that the strong progress theorem from the Smallstep chapter fails here. That is, there are terms that are normal forms (they can't take a step) but not values (they are not included in our definition of possible "results of reduction"). Such terms are said to be stuck.
Notation step_normal_form := (normal_form step).

Definition stuck (t : tm) : Prop :=
step_normal_form t ¬ value t.

Hint Unfold stuck : core.

Exercise: 2 stars, standard (some_term_is_stuck)

Example some_term_is_stuck :
t, stuck t.
Proof.
(* FILL IN HERE *) Admitted.
However, although values and normal forms are not the same in this language, the set of values is a subset of the set of normal forms. This is important because it shows we did not accidentally define things so that some value could still take a step.

Exercise: 3 stars, standard (value_is_nf)

Lemma value_is_nf : t,
value t step_normal_form t.
Proof.
(* FILL IN HERE *) Admitted.

(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, standard, optional (step_deterministic)

Use value_is_nf to 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 is that, although this language has stuck terms, they are always 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 :=
| Bool : ty
| Nat : ty.
In informal notation, the typing relation is often written t \in T and pronounced "t has type T." The symbol is called a "turnstile." Below, we're going to see richer typing relations where one or more additional "context" arguments are written to the left of the turnstile. For the moment, 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_0) ⊢ 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 :
<{ true }> \in Bool
| T_False :
<{ false }> \in Bool
| T_If : t1 t2 t3 T,
t1 \in Bool
t2 \in T
t3 \in T
<{ if t1 then t2 else t3 }> \in T
| T_0 :
<{ 0 }> \in Nat
| T_Succ : t1,
t1 \in Nat
<{ succ t1 }> \in Nat
| T_Pred : t1,
t1 \in Nat
<{ pred t1 }> \in Nat
| T_Iszero : t1,
t1 \in Nat
<{ iszero t1 }> \in Bool

where "'⊢' t '∈' T" := (has_type t T).

Hint Constructors has_type : core.

Example has_type_1 :
<{ if false then 0 else (succ 0) }> \in Nat.
Proof.
apply T_If.
- apply T_False.
- apply T_0.
- apply T_Succ. apply T_0.
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.)
It's important to realize that the typing relation is a conservative (or static) approximation: it does not consider what happens when the term is reduced -- in particular, it does not calculate the type of its normal form.
Example has_type_not :
¬ ( <{ if false then 0 else true}> \in Bool ).
Proof.
intros Contra. solve_by_inverts 2. Qed.

Exercise: 1 star, standard, optional (succ_hastype_nat__hastype_nat)

Example succ_hastype_nat__hastype_nat : t,
<{succ t}> \in Nat
t \in Nat.
Proof.
(* FILL IN HERE *) Admitted.

Canonical forms

The following two lemmas capture the fundamental property that the definitions of boolean and numeric values agree with the typing relation.
Lemma bool_canonical : t,
t \in Bool value t bvalue t.
Proof.
intros t HT [Hb | Hn].
- assumption.
- destruct Hn as [ | Hs].
+ inversion HT.
+ inversion HT.
Qed.

Lemma nat_canonical : t,
t \in Nat value t nvalue t.
Proof.
intros t HT [Hb | Hn].
- inversion Hb; subst; inversion HT.
- assumption.
Qed.

Progress

The typing relation enjoys two critical properties. The first is that well-typed normal forms are not stuck -- or conversely, if a term is well typed, then either it is a value or it can take at least one step. We call this progress.

Exercise: 3 stars, standard (finish_progress)

Theorem progress : t T,
t \in T
value t t', t --> t'.
Complete the formal proof of the progress property. (Make sure you understand the parts we've given of the informal proof in the following exercise before starting -- this will save you a lot of time.)
Proof.
intros t T HT.
induction HT; auto.
(* The cases that were obviously values, like T_True and
T_False, are eliminated immediately by auto *)

- (* T_If *)
right. destruct IHHT1.
+ (* t1 is a value *)
apply (bool_canonical t1 HT1) in H.
destruct H.
× t2. auto.
× t3. auto.
+ (* t1 can take a step *)
destruct H as [t1' H1].
(<{ if t1' then t2 else t3 }>). auto.
(* FILL IN HERE *) Admitted.

Exercise: 3 stars, advanced (finish_progress_informal)

Complete the corresponding informal proof:
Theorem: If t \in T, then either t is a value or else t --> t' for some t'.
Proof: By induction on a derivation of t \in T.
• If the last rule in the derivation is T_If, then t = if t1 then t2 else t3, with t1 \in Bool, t2 \in T and t3 \in 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 \in 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.
• (* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_finish_progress_informal : option (nat×string) := None.
This theorem 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.

Type Preservation

The second critical property of typing is that, when a well-typed term takes a step, the result is a well-typed term (of the same type).

Exercise: 2 stars, standard (finish_preservation)

Theorem preservation : t t' T,
t \in T
t --> t'
t' \in T.
Complete the formal proof of the preservation property. (Again, make sure you understand the informal proof fragment in the following exercise first.)
Proof.
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_invert.
- (* 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.

Exercise: 3 stars, advanced (finish_preservation_informal)

Complete the following informal proof:
Theorem: If t \in T and t --> t', then t' \in T.
Proof: By induction on a derivation of t \in T.
• If the last rule in the derivation is T_If, then t = if t1 then t2 else t3, with t1 \in Bool, t2 \in T and t3 \in 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 \in T, so we are done.
• If the last rule was ST_IfFalse, then t' = t3. But we know that t3 \in 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 \in Bool so, by the IH, t1' \in Bool. The T_If rule then gives us if t1' then t2 else t3 \in T, as required.
• (* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_finish_preservation_informal : option (nat×string) := None.

Exercise: 3 stars, standard (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 proofs 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 \in T
t --> t'
t' \in T.
Proof with eauto.
(* FILL IN HERE *) Admitted.
The preservation theorem is often called subject reduction, 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.

Type Soundness

Putting progress and preservation together, we 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 : t t' T,
t \in T
t -->* t'
~(stuck t').
Proof.
intros t t' T HT P. induction P; intros [R S].
- apply progress in HT. destruct HT; auto.
- apply IHP.
+ apply preservation with (t := x); auto.
+ unfold stuck. split; auto.
Qed.

Exercise: 3 stars, standard, especially useful (subject_expansion)

Having seen the subject reduction property, one might wonder whether the opposite property -- subject expansion -- also holds. That is, is it always the case that, if t --> t' and t' \in T, then t \in T? If so, prove it. If not, give a counter-example.
(* FILL IN HERE *)
Theorem subject_expansion:
( t t' T, t --> t' t' \in T t \in T)

¬ ( t t' T, t --> t' t' \in T t \in T).
Proof.
(* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard (variation1)

Suppose that we add this new rule to the typing relation:
| T_SuccBool : t,
⊢ t \in Bool
⊢ <{ succ t }> \in Bool
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.
• Determinism of step (* FILL IN HERE *)
• Progress (* FILL IN HERE *)
• Preservation (* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_variation1 : option (nat×string) := None.

Exercise: 2 stars, standard (variation2)

Suppose, instead, that we add this new rule to the step relation:
| ST_Funny1 : t2 t3,
(<{ if true then t2 else t3 }>) --> t3
Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example. (* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_variation2 : option (nat×string) := None.

Exercise: 2 stars, standard, optional (variation3)

Suppose instead that we add this rule:
| ST_Funny2 : t1 t2 t2' t3,
t2 --> t2'
(<{ if t1 then t2 else t3 }>) --> (<{ if t1 then t2' else t3 }>)
Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example. (* FILL IN HERE *)

Exercise: 2 stars, standard, optional (variation4)

Suppose instead that we add this rule:
| ST_Funny3 :
(<{pred false}>) --> (<{ pred (pred false)}>)
Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example. (* FILL IN HERE *)

Exercise: 2 stars, standard, optional (variation5)

Suppose instead that we add this rule:
| T_Funny4 :
⊢ <{ 0 }> \in Bool
Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example. (* FILL IN HERE *)

Exercise: 2 stars, standard, optional (variation6)

Suppose instead that we add this rule:
| T_Funny5 :
⊢ <{ pred 0 }> \in Bool
Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example. (* FILL IN HERE *)

Exercise: 3 stars, standard, 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.
(* FILL IN HERE *)

Exercise: 1 star, standard (remove_pred0)

The reduction rule ST_Pred0 is a bit counter-intuitive: we might feel that it makes more sense for the predecessor of <{0}> to be undefined, rather than being defined to be <{0}>. Can we achieve this simply by removing the rule from the definition of step? Would doing so create any problems elsewhere?
(* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_remove_pred0 : option (nat×string) := None.

Exercise: 4 stars, advanced (prog_pres_bigstep)

Suppose our evaluation relation is defined in the big-step style. State appropriate analogs of the progress and preservation properties. (You do not need to prove them.)
Can you see any limitations of either of your properties? Do they allow for nonterminating programs? Why might we prefer the small-step semantics for stating preservation and progress?
(* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_prog_pres_bigstep : option (nat×string) := None.
End TM.

(* 2022-09-20 16:43 *)