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!).

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: And here it is formally:

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

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_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'

... and then 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 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.

Exercise: 2 stars (some_term_is_stuck)

(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)

Use value_is_nf to show that the step relation is also deterministic.

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.

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_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

(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.

Exercise: 1 star, optional (succ_hastype_nat__hastype_nat)

Canonical forms

The following two lemmas capture the fundamental property that the definitions of boolean and numeric values agree with the typing relation.

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 (finish_progress)

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.)

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 (finish_preservation)

Complete the formal proof of the preservation property. (Again, make sure you understand the informal proof fragment in the following exercise first.)

Type Soundness

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

Aside: the normalize Tactic

When experimenting with definitions of programming languages in Coq, we often want to see what a particular concrete term steps to — i.e., we want to find proofs for goals of the form t ==>* t', where t is a completely concrete term and t' is unknown. These proofs are quite tedious to do by hand. Consider, for example, reducing an arithmetic expression using the small-step relation astep.

The proof repeatedly applies multi_step until the term reaches a normal form. Fortunately The sub-proofs for the intermediate steps are simple enough that auto, with appropriate hints, can solve them.

The following custom Tactic Notation definition captures this pattern. In addition, before each step, we print out the current goal, so that we can follow how the term is being reduced.

The normalize tactic also provides a simple way to calculate the normal form of a term, by starting with a goal with an existentially bound variable.

Exercise: 1 star (normalize_ex)

Additional Exercises

Exercise: 2 stars, 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 *)
☐

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 reduction rule ST_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. 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 *)
☐