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".
From Coq Require Import Arith.Arith.
From PLF Require Import Maps.
From PLF Require Import Imp.
From PLF Require Import Smallstep.

Hint Constructors multi.

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 ::= tru
        | fls
        | test t then t else t
        | zro
        | scc t
        | prd t
        | iszro t
And here it is formally:
Inductive tm : Type :=
  | tru : tm
  | fls : tm
  | test : tmtmtmtm
  | zro : tm
  | scc : tmtm
  | prd : tmtm
  | iszro : tmtm.
Values are tru, fls, and numeric values...
Inductive bvalue : tmProp :=
  | bv_tru : bvalue tru
  | bv_fls : bvalue fls.

Inductive nvalue : tmProp :=
  | nv_zro : nvalue zro
  | nv_scc : t, nvalue tnvalue (scc t).

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

Hint Constructors bvalue nvalue.
Hint Unfold value.
Hint Unfold update.

Operational Semantics

Here is the single-step relation, first informally...
   (ST_TestTru)  

test tru then t1 else t2 --> t1
   (ST_TestFls)  

test fls then t1 else t2 --> t2
t1 --> t1' (ST_Test)  

test t1 then t2 else t3 --> test t1' then t2 else t3
t1 --> t1' (ST_Scc)  

scc t1 --> scc t1'
   (ST_PrdZro)  

prd zro --> zro
numeric value v1 (ST_PrdScc)  

prd (scc v1--> v1
t1 --> t1' (ST_Prd)  

prd t1 --> prd t1'
   (ST_IszroZro)  

iszro zro --> tru
numeric value v1 (ST_IszroScc)  

iszro (scc v1--> fls
t1 --> t1' (ST_Iszro)  

iszro t1 --> iszro t1'
... and then formally:
Reserved Notation "t1 '-->' t2" (at level 40).

Inductive step : tmtmProp :=
  | ST_TestTru : t1 t2,
      (test tru t1 t2) --> t1
  | ST_TestFls : t1 t2,
      (test fls t1 t2) --> t2
  | ST_Test : t1 t1' t2 t3,
      t1 --> t1'
      (test t1 t2 t3) --> (test t1' t2 t3)
  | ST_Scc : t1 t1',
      t1 --> t1'
      (scc t1) --> (scc t1')
  | ST_PrdZro :
      (prd zro) --> zro
  | ST_PrdScc : t1,
      nvalue t1
      (prd (scc t1)) --> t1
  | ST_Prd : t1 t1',
      t1 --> t1'
      (prd t1) --> (prd t1')
  | ST_IszroZro :
      (iszro zro) --> tru
  | ST_IszroScc : t1,
       nvalue t1
      (iszro (scc t1)) --> fls
  | ST_Iszro : t1 t1',
      t1 --> t1'
      (iszro t1) --> (iszro t1')

where "t1 '-->' t2" := (step t1 t2).

Hint Constructors step.
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 scc tru cannot take a step, but the almost as obviously nonsensical term
       scc (test tru then tru else tru)
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 (because we have not included them in our definition of possible "results of reduction"). 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, 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 tstep_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 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_Tru)  

⊢ tru ∈ Bool
   (T_Fls)  

⊢ fls ∈ Bool
⊢ t1 ∈ Bool    ⊢ t2 ∈ T    ⊢ t3 ∈ T (T_Test)  

⊢ test t1 then t2 else t3 ∈ T
   (T_Zro)  

⊢ zro ∈ Nat
⊢ t1 ∈ Nat (T_Scc)  

⊢ scc t1 ∈ Nat
⊢ t1 ∈ Nat (T_Prd)  

⊢ prd t1 ∈ Nat
⊢ t1 ∈ Nat (T_IsZro)  

⊢ iszro t1 ∈ Bool
Reserved Notation "'⊢' t '∈' T" (at level 40).

Inductive has_type : tmtyProp :=
  | T_Tru :
       ⊢ truBool
  | T_Fls :
       ⊢ flsBool
  | T_Test : t1 t2 t3 T,
       ⊢ t1Bool
       ⊢ t2T
       ⊢ t3T
       ⊢ test t1 t2 t3T
  | T_Zro :
       ⊢ zroNat
  | T_Scc : t1,
       ⊢ t1Nat
       ⊢ scc t1Nat
  | T_Prd : t1,
       ⊢ t1Nat
       ⊢ prd t1Nat
  | T_Iszro : t1,
       ⊢ t1Nat
       ⊢ iszro t1Bool

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

Hint Constructors has_type.

Example has_type_1 :
  ⊢ test fls zro (scc zro) ∈ Nat.
Proof.
  apply T_Test.
    - apply T_Fls.
    - apply T_Zro.
    - apply T_Scc.
       + apply T_Zro.
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 :
  ¬( ⊢ test fls zro truBool ).
Proof.
  intros Contra. solve_by_inverts 2. Qed.

Exercise: 1 star, standard, optional (scc_hastype_nat__hastype_nat)

Example scc_hastype_nat__hastype_nat : t,
  ⊢ scc tNat
  ⊢ tNat.
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,
  ⊢ tBoolvalue tbvalue t.
Proof.
  intros t HT [Hb | Hn].
  - assumption.
  - induction Hn; inversion HT; auto.
Qed.

Lemma nat_canonical : t,
  ⊢ tNatvalue tnvalue 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,
  ⊢ tT
  value tt', 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 with auto.
  intros t T HT.
  induction HT...
  (* The cases that were obviously values, like T_Tru and
     T_Fls, were eliminated immediately by auto *)

  - (* T_Test *)
    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].
      (test t1' t2 t3)...
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars, advanced (finish_progress_informal)

Complete the corresponding informal 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_Test, then t = test 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 tru or fls. If t1 = tru, then t steps to t2 by ST_TestTru, while if t1 = fls, then t steps to t3 by ST_TestFls. Either way, t can step, which is what we wanted to show.
    • If t1 itself can take a step, then, by ST_Test, 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 also a well-typed term.

Exercise: 2 stars, standard (finish_preservation)

Theorem preservation : t t' T,
  ⊢ tT
  t --> t'
  ⊢ t'T.
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_invert.
    - (* T_Test *) inversion HE; subst; clear HE.
      + (* ST_TESTTru *) assumption.
      + (* ST_TestFls *) assumption.
      + (* ST_Test *) apply T_Test; try assumption.
        apply IHHT1; assumption.
    (* FILL IN HERE *) Admitted.

Exercise: 3 stars, advanced (finish_preservation_informal)

Complete the following informal 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_Test, then t = test 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 test ..., we see that the only ones that could have been used to prove t --> t' are ST_TestTru, ST_TestFls, or ST_Test.
    • If the last rule was ST_TestTru, then t' = t2. But we know that t2 T, so we are done.
    • If the last rule was ST_TestFls, then t' = t3. But we know that t3 T, so we are done.
    • If the last rule was ST_Test, then t' = test t1' then t2 else t3, where t1 --> t1'. We know t1 Bool so, by the IH, t1' Bool. The T_Test rule then gives us test t1' then t2 else t3 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,
  ⊢ tT
  t --> t'
  ⊢ t'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,
  ⊢ tT
  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, standard, recommended (subject_expansion)

Having seen the subject reduction property, one might 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.)
(* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_subject_expansion : option (nat*string) := None.

Exercise: 2 stars, standard (variation1)

Suppose that we add this new rule to the typing relation:
      | T_SccBool : t,
           ⊢ t ∈ Bool →
           ⊢ scc t ∈ 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,
           (test tru 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. (* 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' →
           (test t1 t2 t3--> (test 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. (* FILL IN HERE *)

Exercise: 2 stars, standard, optional (variation4)

Suppose instead that we add this rule:
      | ST_Funny3 :
          (prd fls--> (prd (prd fls))
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 :
            ⊢ zro ∈ 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 :
            ⊢ prd zro ∈ 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_prdzro)

The reduction rule ST_PrdZro is a bit counter-intuitive: we might feel that it makes more sense for the predecessor of zro to be undefined, rather than being defined to be zro. 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_predzro : 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 commands? 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.
(* Wed Jan 9 12:05:11 EST 2019 *)