Our next topic, a large one, 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).

More Automation

Before we start, let's spend a little time learning to use some of Coq's more powerful automation features...

The auto and eauto Tactics

The auto tactic solves goals that are solvable by any combination of

• intros,
• apply (with a local hypothesis, by default), and
• reflexivity.

The eauto tactic works just like auto, except that it uses eapply instead of apply. Using auto is always "safe" in the sense that it will never fail and will never change the proof state: either it completely solves the current goal, or it does nothing. Here is a contrived example:

When searching for potential proofs of the current goal, auto and eauto consider the hypotheses in the current context together with a hint database of other lemmas and constructors. Some of the lemmas and constructors we've already seen — e.g., conj, or_introl, and or_intror — are installed in this hint database by default.

We can extend the hint database just for the purposes of one application of auto or eauto by writing auto using .... E.g., if conj, or_introl, and or_intror had not already been in the hint database, we could have done this instead:

Of course, in any given development there will also be some of our own specific constructors and lemmas that are used very often in proofs. We can add these to the global hint database by writing at the top level, where l is a top-level lemma theorem or a constructor of an inductively defined proposition (i.e., anything whose type is an implication). As a shorthand, we can write to tell Coq to do a Hint Resolve for all of the constructors from the inductive definition of c. It is also sometimes necessary to add where d is a defined symbol, so that auto knows to expand uses of d and enable further possibilities for applying lemmas that it knows about. Here are some Hints we will find useful.

Warning: Just as with Coq's other automation facilities, it is easy to overuse auto and eauto and wind up with proofs that are impossible to understand later! Also, overuse of eauto can make proof scripts very slow. Get in the habit of using auto most of the time and eauto only when necessary. For much more detailed information about using auto and eauto, see the chapter UseAuto.v.

The Proof with Tactic

If you start a proof by saying Proof with (tactic) instead of just Proof, then writing ... instead of . after a tactic in the body of the proof will try to solve all generated subgoals with tactic (and fail if this doesn't work). One common use of this facility is "Proof with auto" (or eauto). We'll see many examples of this later in the file.

The solve by inversion Tactic

Here's another nice automation feature: it often arises that the context contains a contradictory assumption and we want to use inversion on it to solve the goal. We'd like to be able to say to Coq, "find a contradictory assumption and invert it" without giving its name explicitly. Doing solve by inversion will find a hypothesis that can be inverted to solve the goal, if there is one. The tactics solve by inversion 2 and solve by inversion 3 are slightly fancier versions which will perform two or three inversions in a row, if necessary, to solve the goal. (These tactics are not actually built into Coq — their definitions are in Sflib.v.) Caution: Overuse of solve by inversion can lead to slow proof scripts.

The try solve Tactic

If t is a tactic, then try solve [t] is a tactic that

• if t solves the goal, behaves just like t, or
• if t cannot completely solve the goal, does nothing.

More generally, try solve [t1 | t2 | ...] will try to solve the goal by using t1, t2, etc. If none of them succeeds in completely solving the goal, then try solve [t1 | t2 | ...] does nothing. The fact that it does nothing when it doesn't completely succeed in solving the goal means that there is no harm in using try solve T in situations where T might actually make no sense. In particular, if we put it after a ; it will solve as many subgoals as it can and leave the rest for us to solve by other methods. It will not leave any of the subgoals in a partially solved state.

The f_equal Tactic

This handy tactic replaces a goal of the form f x1 x2 ... xn = f y1 y2 ... yn, where f is some function, with the subgoals x1 = y1, x2 = y2,...,xn = yn. It is useful for avoiding explicit rewriting steps, and often the generated subgoals can be quickly cleared by auto.

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 simple but repetitive to do by hand. Consider for example reducing an arithmetic expression using the small-step relation astep defined in the previous chapter:

We repeatedly applied rsc_step until we got to a normal form. The proofs that the intermediate steps are possible are simple enough that auto, with appropriate hints, can solve them.

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

Finally, this is also useful as a simple way to calculate what the normal form of a term is, by proving a goal with an existential variable in it.

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 Smallstep.v: 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 ImpList.v 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

Small-Step Reduction

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 tm_succ tm_true cannot take a step, but the almost-as-obviously-nonsensical term can.

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, optional (some_tm_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, optional (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.

The typing relation: 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 (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, recommended (finish_progress_informal)

Complete the following 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 *)
☐

Exercise: 3 stars (finish_progress)

☐ 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. (As usual, no need to hand these in.)

• 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.
• 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: 3 stars, recommended (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: 2 stars (finish_preservation)

☐

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.

Indeed, in the present — extremely simple — language, every well-typed term can be reduced to a normal form: this is the normalization property. (The proof is straightforward, though somewhat tedious.) In richer languages, this property often fails, though there are some interesting languages (such as Coq's Fixpoint language, and the simply typed lambda-calculus, which we'll be looking at next) where all well-typed terms can be reduced to normal forms.

Additional Exercises

Exercise: 2 stars, recommended (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 has_type t' T, then has_type t T? If so, prove it. If not, give a counter-example. (* FILL IN HERE *)
☐

Exercise: 2 stars (variation1)

Suppose we add the following two new rules to the evaluation relation: Which of the following properties remain true in the presence of these rules? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample.

• Determinacy of step
• Normalization of step for well-typed terms
• Progress
• Preservation

☐

Exercise: 2 stars (variation2)

Suppose, instead, that we add this new rule to the typing relation: Which of the following properties remain true in the presence of this rule? (Answer in the same style as above.)

• Determinacy of step
• Normalization of step for well-typed terms
• Progress
• Preservation

☐

Exercise: 2 stars (variation3)

Suppose, instead, that we add this new rule to the typing relation: Which of the following properties remain true in the presence of this rule? (Answer in the same style as above.)

• Determinacy of step
• Normalization of step for well-typed terms
• Progress
• Preservation

☐

Exercise: 2 stars (variation4)

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 (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 (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: 2 stars (variation7)

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 (variation8)

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? (* FILL IN HERE *)
☐

Exercise: 4 stars, optional (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 *)
☐

The Simply Typed Lambda-Calculus

The simply typed lambda-calculus (STLC) is a tiny core calculus embodying the key concept of functional abstraction, which shows up in pretty much every real-world programming language in some form (functions, procedures, methods, etc.). We will follow exactly the same pattern as above when formalizing of this calculus (syntax, small-step semantics, typing rules) and its main properties (progress and preservation). The new technical challenges (which will take some work to deal with) all arise from the mechanisms of variable binding and substitution.

Overview

The STLC is built on some collection of base types — booleans, numbers, strings, etc. The exact choice of base types doesn't matter — the construction of the language and its theoretical properties work out pretty much the same — so for the sake of brevity let's take just Bool for the moment. At the end of the chapter we'll see how to add more base types, and in later chapters we'll enrich the pure STLC with other useful constructs like pairs, records, subtyping, and mutable state. Starting from the booleans, we add three things:

• variables
• function abstractions
• application

This gives us the following collection of abstract syntax constructors (written out here in informal BNF notation — we'll formalize it below):

<
       t ::= x                       variable
           | \x:T.t1                 abstraction
           | t1 t2                   application
           | true                    constant true
           | false                   constant false
           | if t1 then t2 else t3   conditional

The \ symbol in a function abstraction \x:T.t1 is often written as a greek "lambda" (hence the name of the calculus). The variable x is called the parameter to the function; the term t1 is its body. The annotation :T specifies the type of arguments that the function can be applied to. Some examples:

• \x:Bool. x
  The identity function for booleans.
• (\x:Bool. x) true
  The identity function for booleans, applied to the boolean true.
• \x:Bool. if x then false else true
  The boolean "not" function.
• \x:Bool. true
  The constant function that takes every (boolean) argument to true.
• \x:Bool. \y:Bool. x
  A two-argument function that takes two booleans and returns the first one. (Note that, as in Coq, a two-argument function is really a one-argument function whose body is also a one-argument function.)
• (\x:Bool. \y:Bool. x) false true
  A two-argument function that takes two booleans and returns the first one, applied to the booleans false and true. Note that, as in Coq, application associates to the left — i.e., this expression is parsed as ((\x:Bool. \y:Bool. x) false) true.
• \f:Bool→Bool. f (f true)
  A higher-order function that takes a function f (from booleans to booleans) as an argument, applies f to true, and applies f again to the result.
• (\f:Bool→Bool. f (f true)) (\x:Bool. false)
  The same higher-order function, applied to the constantly false function.

As the last several examples show, the STLC is a language of higher-order functions: we can write down functions that take other functions as arguments and/or return other functions as results. Another point to note is that the STLC doesn't provide any primitive syntax for defining named functions — all functions are "anonymous." We'll see in chapter MoreStlc that it is easy to add named functions to what we've got — indeed, the fundamental naming and binding mechanisms are exactly the same. The types of the STLC include Bool, which classifies the boolean constants true and false as well as more complex computations that yield booleans, plus arrow types that classify functions.

      T ::= Bool
          | T1 -> T2

For example:

• \x:Bool. false has type Bool→Bool
• \x:Bool. x has type Bool→Bool
• (\x:Bool. x) true has type Bool
• \x:Bool. \y:Bool. x has type Bool→Bool→Bool (i.e. Bool → (Bool→Bool))
• (\x:Bool. \y:Bool. x) false has type Bool→Bool
• (\x:Bool. \y:Bool. x) false true has type Bool
• \f:Bool→Bool. f (f true) has type (Bool→Bool) → Bool
• (\f:Bool→Bool. f (f true)) (\x:Bool. false) has type Bool

Syntax

Types

Terms

Something to note here is that an abstraction \x:T.t (formally, tm_abs x T t) is always annotated with the type (T) of its parameter. This is in contrast to Coq (and other functional languages like ML, Haskell, etc.), which use type inference to fill in missing annotations.

Some examples...

idB = \a:Bool. a

idBB = \a:Bool→Bool. a

idBBBB = \a:(Bool→Bool)->(Bool→Bool). a

k = \a:Bool. \b:Bool. a

(We write these as Notations rather than Definitions to make things easier for auto.)

Operational Semantics

To define the small-step semantics of STLC terms, we begin — as always — by defining the set of values. Next, we define the critical notions of free variables and substitution, which are used in the reduction rule for application expressions. And finally we give the small-step relation itself.

Values

To define the values of the STLC, we have a few cases to consider. First, for the boolean part of the language, the situation is clear: true and false are the only values. (An if expression is never a value.) Second, an application is clearly not a value: It represents a function being invoked on some argument, which clearly still has work left to do. Third, for abstractions, we have a choice:

• We can say that \a:A.t1 is a value only when t1 is a value — i.e., only if the function's body has been reduced (as much as it can be without knowing what argument it is going to be applied to).
• Or we can say that \a:A.t1 is always a value, no matter whether t1 is one or not — in other words, we can say that reduction stops at abstractions.

Coq makes the first choice — for example, yields fun a:bool => 7. But most real functional programming languages make the second choice — reduction of a function's body only begins when the function is actually applied to an argument. We also make the second choice here. Finally, having made the choice not to reduce under abstractions, we don't need to worry about whether variables are values, since we'll always be reducing programs "from the outside in," and that means the step relation will always be working with closed terms (ones with no free variables).

Free Variables and Substitution

Now we come to the heart of the matter: the operation of substituting one term for a variable in another term. This operation will be used below to define the operational semantics of function application, where we will need to substitute the argument term for the function parameter in the function's body. For example, we reduce to false by substituting false for the parameter x in the body of the function. In general, we need to be able to substitute some given term s for occurrences of some variable x in another term t. In informal discussions, this is usually written [s/x]t and pronounced "substitute s for x in t." Here are some examples:

• [true / a] (if a then a else false) yields if true then true else false
• [true / a] a yields true
• [true / a] (if a then a else b) yields if true then true else b
• [true / a] b yields b
• [true / a] false yields false (vacuous substitution)
• [true / a] (\y:Bool. if y then a else false) yields \y:Bool. if y then true else false
• [true / a] (\y:Bool. a) yields \y:Bool. true
• [true / a] (\y:Bool. y) yields \y:Bool. y
• [true / a] (\a:Bool. a) yields \a:Bool. a

The last example is very important: substituting true for a in \a:Bool. a does not yield \a:Bool. true! The reason for this is that the a in the body of \a:Bool. a is bound by the abstraction: it is a new, local name that just happens to be spelled the same as some global name a. Here is the definition, informally... ... and formally:

Technical note: Substitution becomes trickier to define if we consider the case where s, the term being substituted for a variable in some other term, may itself contain free variables. Since we are only interested in defining the step relation on closed terms here, we can avoid this extra complexity.

Reduction

The small-step reduction relation for STLC follows the same pattern as the ones we have seen before. Intuitively, to reduce a function application, we first reduce its left-hand side until it becomes a literal function; then we reduce its right-hand side (the argument) until it is also a value; and finally we substitute the argument for the bound variable in the body of the function. This last rule, written informally as is traditionally called "beta-reduction". Informally:

	(ST_AppAbs)
(\a:T.t12) v2 ⇒ [v2/a]t12

t1 ⇒ t1'	(ST_App1)
t1 t2 ⇒ t1' t2

t2 ⇒ t2'	(ST_App2)
v1 t2 ⇒ v1 t2'

(plus the usual rules for booleans). Formally:

Examples

Again, we can use the normalize tactic from above to simplify the proof.

Exercise: 2 stars (step_example3)

Try to do this one both with and without normalize.

☐

Typing

Contexts

Question: What is the type of the term "x y"? Answer: It depends on the types of x and y! I.e., in order to assign a type to a term, we need to know what assumptions we should make about the types of its free variables. This leads us to a three-place "typing judgment", informally written Γ ⊢ t : T, where Γ is a "typing context" a mapping from variables to their types. We hide the definition of partial maps in a module since it is actually defined in SfLib.

Typing Relation

Informally:

Γ x = T	(T_Var)
Γ ⊢ x : T

Γ , x:T11 ⊢ t12 : T12	(T_Abs)
Γ ⊢ \x:T11.t12 : T11->T12

Γ ⊢ t1 : T11->T12
Γ ⊢ t2 : T11	(T_App)
Γ ⊢ t1 t2 : T12

	(T_True)
Γ ⊢ true : Bool

	(T_False)
Γ ⊢ false : Bool

Γ ⊢ t1 : Bool    Γ ⊢ t2 : T    Γ ⊢ t3 : T	(T_If)
Γ ⊢ if t1 then t2 else t3 : T

The notation Γ , x:T means "extend the partial function Γ to also map x to T."

Examples

Note that since we added the has_type constructors to the hints database, auto can actually solve this one immediately.

Written informally, the next one is:

Exercise: 2 stars, optional

Prove the same result without using auto, eauto, or eapply.

☐

Exercise: 2 stars (typing_example_3)

Formally prove the following typing derivation holds:

☐ We can also show that terms are not typable. For example, let's formally check that there is no typing derivation assigning a type to the term \a:Bool. \b:Bool, a b — i.e.,

Exercise: 3 stars (typing_nonexample_3)

Another nonexample:

☐

Exercise: 1 star (typing_statements)

Which of the following propositions are provable?

• b:Bool ⊢ \a:Bool.a : Bool→Bool
• ∃ T, empty ⊢ (\b:Bool→Bool. \a:Bool. b a) : T
• ∃ T, empty ⊢ (\b:Bool→Bool. \a:Bool. a b) : T
• ∃ S, a:S ⊢ (\b:Bool→Bool. b) a : S
• ∃ S, ∃ T, a:S ⊢ (a a a) : T

☐

Exercise: 1 star, optional (more_typing_statements)

Which of the following propositions are provable? For the ones that are, give witnesses for the existentially bound variables.

• ∃ T, empty ⊢ (\b:B→B→B. \a:B, b a) : T
• ∃ T, empty ⊢ (\a:A→B, \b:B-->C, \c:A, b (a c)):T
• ∃ S, ∃ U, ∃ T, a:S, b:U ⊢ \c:A. a (b c) : T
• ∃ S, ∃ T, a:S ⊢ \b:A. a (a b) : T
• ∃ S, ∃ U, ∃ T, a:S ⊢ a (\c:U. c a) : T

☐

Properties

Free Occurrences

A variable x appears free in a term t if t contains some occurrence of x that is not under an abstraction labeled x. For example:

• y appears free, but x does not, in \x:T→U. x y
• both x and y appear free in (\x:T→U. x y) x
• no variables appear free in \x:T→U. \y:T. x y

A term in which no variables appear free is said to be closed.

Substitution

We first need a technical lemma connecting free variables and typing contexts. If a variable x appears free in a term t, and if we know t is well typed in context Γ, then it must be the case that Γ assigns a type to x.

Proof: We show, by induction on the proof that x appears free in t, that, for all contexts Γ, if t is well typed under Γ, then Γ assigns some type to x.

• If the last rule used was afi_var, then t = x, and from the assumption that t is well typed under Γ we have immediately that Γ assigns a type to x.
• If the last rule used was afi_app1, then t = t1 t2 and x appears free in t1. Since t is well typed under Γ, we can see from the typing rules that t1 must also be, and the IH then tells us that Γ assigns x a type.
• Almost all the other cases are similar: x appears free in a subterm of t, and since t is well typed under Γ, we know the subterm of t in which x appears is well typed under Γ as well, and the IH gives us exactly the conclusion we want.
• The only remaining case is afi_abs. In this case t = \y:T11.t12, and x appears free in t12; we also know that x is different from y. The difference from the previous cases is that whereas t is well typed under Γ, its body t12 is well typed under (Γ, y:T11), so the IH allows us to conclude that x is assigned some type by the extended context (Γ, y:T11). To conclude that Γ assigns a type to x, we appeal to lemma extend_neq, noting that x and y are different variables.

Next, we'll need the fact that any term t which is well typed in the empty context is closed — that is, it has no free variables.

Exercise: 2 stars (typable_empty__closed)

☐ Sometimes, when we have a proof Γ ⊢ t : T, we will need to replace Γ by a different context Gamma'. When is it safe to do this? Intuitively, it must at least be the case that Gamma' assigns the same types as Γ to all the variables that appear free in t. In fact, this is the only condition that is needed.

Proof: By induction on a derivation of Γ ⊢ t : T.

• If the last rule in the derivation was T_Var, then t = x and Γ x = T. By assumption, Gamma' x = T as well, and hence Gamma' ⊢ t : T by T_Var.
• If the last rule was T_Abs, then t = \y:T11. t12, with T = T11 → T12 and Γ, y:T11 ⊢ t12 : T12. The induction hypothesis is that for any context Gamma'', if Γ, y:T11 and Gamma'' assign the same types to all the free variables in t12, then t12 has type T12 under Gamma''. Let Gamma' be a context which agrees with Γ on the free variables in t; we must show Gamma' ⊢ \y:T11. t12 : T11 → T12.
  By T_Abs, it suffices to show that Gamma', y:T11 ⊢ t12 : T12. By the IH (setting Gamma'' = Gamma', y:T11), it suffices to show that Γ, y:T11 and Gamma', y:T11 agree on all the variables that appear free in t12.
  Any variable occurring free in t12 must either be y, or some other variable. Γ, y:T11 and Gamma', y:T11 clearly agree on y. Otherwise, we note that any variable other than y which occurs free in t12 also occurs free in t = \y:T11. t12, and by assumption Γ and Gamma' agree on all such variables, and hence so do Γ, y:T11 and Gamma', y:T11.
• If the last rule was T_App, then t = t1 t2, with Γ ⊢ t1 : T2 → T and Γ ⊢ t2 : T2. One induction hypothesis states that for all contexts Gamma', if Gamma' agrees with Γ on the free variables in t1, then t1 has type T2 → T under Gamma'; there is a similar IH for t2. We must show that t1 t2 also has type T under Gamma', given the assumption that Gamma' agrees with Γ on all the free variables in t1 t2. By T_App, it suffices to show that t1 and t2 each have the same type under Gamma' as under Γ. However, we note that all free variables in t1 are also free in t1 t2, and similarly for free variables in t2; hence the desired result follows by the two IHs.

Now we come to the conceptual heart of the proof that reduction preserves types — namely, the observation that substitution preserves types. Formally, the so-called Substitution Lemma says this: suppose we have a term t with a free variable x, and suppose we've been able to assign a type T to t under the assumption that x has some type U. Also, suppose that we have some other term v and that we've shown that v has type U. Then, since v satisfies the assumption we made about x when typing t, we should be able to substitute v for each of the occurrences of x in t and obtain a new term that still has type T. Lemma: If Γ,x:U ⊢ t : T and ⊢ v : U, then Γ ⊢ [v/x]t : T.

One technical subtlety in the statement of the lemma is that we assign v the type U in the empty context — in other words, we assume v is closed. This assumption considerably simplifies the T_Abs case of the proof (compared to assuming Γ ⊢ v : U, which would be the other reasonable assumption at this point) because the context invariance lemma then tells us that v has type U in any context at all — we don't have to worry about free variables in v clashing with the variable being introduced into the context by T-Abs. Proof: We prove, by induction on t, that, for all T and Γ, if Γ,x:U ⊢ t : T and ⊢ v : U, then Γ ⊢ [v/x]t : T.

• If t is a variable, there are two cases to consider, depending on whether t is x or some other variable.
  • If t = x, then from the fact that Γ, x:U ⊢ x : T we conclude that U = T. We must show that [v/x]x = v has type T under Γ, given the assumption that v has type U = T under the empty context. This follows from context invariance: if a closed term has type T in the empty context, it has that type in any context.
  • If t is some variable y that is not equal to x, then we need only note that y has the same type under Γ, x:U as under Γ.
• If t is an abstraction \y:T11. t12, then the IH tells us, for all Gamma' and T', that if Gamma',x:U ⊢ t12 : T' and ⊢ v : U, then Gamma' ⊢ [v/x]t12 : T'. In particular, if Γ,y:T11,x:U ⊢ t12 : T12 and ⊢ v : U, then Γ,y:T11 ⊢ [v/x]t12 : T12. There are again two cases to consider, depending on whether x and y are the same variable name.
  First, suppose x = y. Then, by the definition of substitution, [v/x]t = t, so we just need to show Γ ⊢ t : T. But we know Γ,x:U ⊢ t : T, and since the variable y does not appear free in \y:T11. t12, the context invariance lemma yields Γ ⊢ t : T.
  Second, suppose x <> y. We know Γ,x:U,y:T11 ⊢ t12 : T12 by inversion of the typing relation, and Γ,y:T11,x:U ⊢ t12 : T12 follows from this by the context invariance lemma, so the IH applies, giving us Γ,y:T11 ⊢ [v/x]t12 : T12. By T_Abs, Γ ⊢ \y:T11. [v/x]t12 : T11→T12, and by the definition of substitution (noting that x <> y), Γ ⊢ \y:T11. [v/x]t12 : T11→T12, as required.
• If t is an application t1 t2, the result follows straightforwardly from the definition of substitution and the induction hypotheses.
• The remaining cases are similar to the application case.

Another technical note: This proof is a rare case where an induction on terms, rather than typing derivations, yields a simpler argument. The reason for this is that the assumption has_type (extend Γ x U) t T is not completely generic, in the sense that one of the "slots" in the typing relation — namely the context — is not just a variable, and this means that Coq's native induction tactic does not give us the induction hypothesis that we want. It is possible to work around this, but the needed generalization is a little tricky. The term t, on the other hand, is completely generic.

The substitution lemma can be viewed as a kind of "commutation" property. Intuitively, it says that substitution and typing can be done in either order: we can either assign types to the terms t and v separately (under suitable contexts) and then combine them using substitution, or we can substitute first and then assign a type to [v/x] t — the result is the same either way.

Preservation

We now have the tools we need to prove preservation: if a closed term t has type T, and takes an evaluation step to t', then t' is also a closed term with type T. In other words, the small-step evaluation relation preserves types.

Proof: by induction on the derivation of ⊢ t : T.

• We can immediately rule out T_Var, T_Abs, T_True, and T_False as the final rules in the derivation, since in each of these cases t cannot take a step.
• If the last rule in the derivation was T_App, then t = t1 t2. There are three cases to consider, one for each rule that could have been used to show that t1 t2 takes a step to t'.
  • If t1 t2 takes a step by ST_App1, with t1 stepping to t1', then by the IH t1' has the same type as t1, and hence t1' t2 has the same type as t1 t2.
  • The ST_App2 case is similar.
  • If t1 t2 takes a step by ST_AppAbs, then t1 = \x:T11.t12 and t1 t2 steps to subst t2 x t12; the desired result now follows from the fact that substitution preserves types.
• If the last rule in the derivation was T_If, then t = if t1 then t2 else t3, and there are again three cases depending on how t steps.
  • If t steps to t2 or t3, the result is immediate, since t2 and t3 have the same type as t.
  • Otherwise, t steps by ST_If, and the desired conclusion follows directly from the induction hypothesis.

Exercise: 2 stars, recommended (subject_expansion_stlc)

An exercise earlier in this file asked about the subject expansion property for the simple language of arithmetic and boolean expressions. Does this property hold for STLC? That is, is it always the case that, if t ⇒ t' and has_type t' T, then has_type t T? If so, prove it. If not, give a counter-example. (* FILL IN HERE *)
☐

Progress

Finally, the progress theorem tells us that closed, well-typed terms are never stuck: either a well-typed term is a value, or else it can take an evaluation step.

Proof: by induction on the derivation of ⊢ t : T.

• The last rule of the derivation cannot be T_Var, since a variable is never well typed in an empty context.
• The T_True, T_False, and T_Abs cases are trivial, since in each of these cases we know immediately that t is a value.
• If the last rule of the derivation was T_App, then t = t1 t2, and we know that t1 and t2 are also well typed in the empty context; in particular, there exists a type T2 such that ⊢ t1 : T2 → T and ⊢ t2 : T2. By the induction hypothesis, either t1 is a value or it can take an evaluation step.
  • If t1 is a value, we now consider t2, which by the other induction hypothesis must also either be a value or take an evaluation step.
    • Suppose t2 is a value. Since t1 is a value with an arrow type, it must be a lambda abstraction; hence t1 t2 can take a step by ST_AppAbs.
    • Otherwise, t2 can take a step, and hence so can t1 t2 by ST_App2.
  • If t1 can take a step, then so can t1 t2 by ST_App1.
• If the last rule of the derivation was T_If, then t = if t1 then t2 else t3, where t1 has type Bool. By the IH, t1 is either a value or takes a step.
  • If t1 is a value, then since it has type Bool it must be either true or false. If it is true, then t steps to t2; otherwise it steps to t3.
  • Otherwise, t1 takes a step, and therefore so does t (by ST_If).

Exercise: 3 stars, optional (progress_from_term_ind)

Show that progress can also be proved by induction on terms instead of types.

☐

Uniqueness of Types

Exercise: 3 stars (types_unique)

Another pleasant property of the STLC is that types are unique: a given term (in a given context) has at most one type. Formalize this statement and prove it.

☐

Additional Exercises

Exercise: 1 star (progress_preservation_statement)

Without peeking, write down the progress and preservation theorems for the simply typed lambda-calculus. ☐

Exercise: 2 stars, optional (stlc_variation1)

Suppose we add the following new rule to the evaluation relation of the STLC: Which of the following properties of the STLC 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.

• Determinacy of step
• Progress
• Preservation

☐

Exercise: 2 stars (stlc_variation2)

Suppose we remove the rule ST_App1 from the step relation. Which of the three properties in the previous exercise become false in the absence of this rule? For each that becomes false, give a counterexample. ☐

Exercise: STLC with Arithmetic

To see how the STLC might function as the core of a real programming language, let's extend it with a concrete base type of numbers and some constants and primitive operators.

Syntax and Operational Semantics

To types, we add a base type of natural numbers (and remove booleans, for brevity)

To terms, we add natural number constants, along with successor, predecessor, multiplication, and zero-testing...