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...

Exercise: 4 stars, recommended (STLCArith)

Finish formalizing the definition and properties of the STLC extended with arithmetic. Specifically:

• Copy the whole development of STLC that we went through above (from the definition of values through the Progress theorem), and paste it into the file at this point.
• Extend the definitions of the subst operation and the step relation to include appropriate clauses for the arithmetic operators.
• Extend the proofs of all the properties of the original STLC to deal with the new syntactic forms. Make sure Coq accepts the whole file.

☐