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

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 HoareList 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: Formally:

Values are true, false, and numeric values...

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:

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 can take a step (once, before becoming stuck).

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.

Exercise: 2 stars (some_term_is_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

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.

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

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.

(Since we've included all the constructors of the typing relation in the hint database, the auto tactic can actually find this proof automatically.)

Exercise: 1 star, optional (succ_hastype_nat__hastype_nat)

☐

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 (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.)

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_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 *)
☐ 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)

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: 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.)

☐

Exercise: 3 stars, advanced (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: 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.

☐

Type Soundness

Putting progress and preservation together, we can see that a well-typed term can never reach a stuck state.

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.) (* FILL IN HERE *)
☐

Exercise: 2 stars (variation1)

Suppose, that we add this new rule to the typing relation: 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
• Progress
• Preservation

☐

Exercise: 2 stars (variation2)

Suppose, instead, that we add this new rule to the step relation: 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, optional (variation3)

Suppose instead that we add this rule: 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, optional (variation4)

Suppose instead that we add this rule: 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, optional (variation5)

Suppose instead that we add this rule: 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, optional (variation6)

Suppose instead that we add this rule: 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, 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? (* FILL IN HERE *)
☐