Stlc: The Simply Typed Lambda-Calculus

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 in the previous chapter when formalizing 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). Informal concrete syntax: t ::= x variable | \x:T1.t2 abstraction | t1 t2 application | true constant true | false constant false | if t1 then t2 else t3 conditional The \ symbol (backslash, in ascii) in a function abstraction \x:T1.t2 is generally written as a greek letter "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

Syntax

Types

Terms

Note that an abstraction \x:T.t (formally, tabs x T t) is always annotated with the type T of its parameter, in contrast to Coq (and other functional languages like ML, Haskell, etc.), which use type inference to fill in missing annotations. We're not considering type inference here, to keep things simple. Some examples...

idB = \x:Bool. x

idBB = \x:Bool→Bool. x

idBBBB = \x:(Bool→Bool) → (Bool→Bool). x

k = \x:Bool. \y:Bool. x

notB = \x:Bool. if x then false else true

(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 \x:T.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 \x:T.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, in its built-in functional programming langauge, makes the first choice -- for example, Eval simpl in (fun x:bool => 3 + 4) yields fun x:bool ⇒ 7. Most real-world 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, we must consider what constitutes a complete program. Intuitively, a "complete" program must not refer to any undefined variables. We'll see shortly how to define the "free" variables in a STLC term. A program is "closed", that is, it contains no free variables. 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).

Substitution

Now we come to the heart of the STLC: 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 (\x:Bool. if x then true else x) false to if false then true else 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 [x:=s]t and pronounced "substitute x with s in t." Here are some examples:

• [x:=true] (if x then x else false) yields if true then true else false
• [x:=true] x yields true
• [x:=true] (if x then x else y) yields if true then true else y
• [x:=true] y yields y
• [x:=true] false yields false (vacuous substitution)
• [x:=true] (\y:Bool. if y then x else false) yields \y:Bool. if y then true else false
• [x:=true] (\y:Bool. x) yields \y:Bool. true
• [x:=true] (\y:Bool. y) yields \y:Bool. y
• [x:=true] (\x:Bool. x) yields \x:Bool. x

The last example is very important: substituting x with true in \x:Bool. x does not yield \x:Bool. true! The reason for this is that the x in the body of \x:Bool. x is bound by the abstraction: it is a new, local name that just happens to be spelled the same as some global name x. Here is the definition, informally... x:=sx = s x:=sy = y if x <> y x:=s(\x:T11.t12) = \x:T11. t12 x:=s(\y:T11.t12) = \y:T11. x:=st12 if x <> y x:=s(t1 t2) = (x:=st1) (x:=st2) x:=strue = true x:=sfalse = false x:=s(if t1 then t2 else t3) = if x:=st1 then x:=st2 else x:=st3 ]] ... 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 here in defining the step relation on closed terms (i.e., terms like \x:Bool. x, that do not mention variables are not bound by some enclosing lambda), we can skip this extra complexity here, but it must be dealt with when formalizing richer languages.

Exercise: 3 stars (substi)

The definition that we gave above uses Coq's Fixpoint facility to define substitution as a function. Suppose, instead, we wanted to define substitution as an inductive relation substi. We've begun the definition by providing the Inductive header and one of the constructors; your job is to fill in the rest of the constructors.

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Reduction

The small-step reduction relation for STLC now 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 (\x:T.t12) v2 ==> x:=v2t12 is traditionally called "beta-reduction". value v2

---

(ST_AppAbs) (\x:T.t12) v2 ==> x:=v2t12 t1 ==> t1'

---

(ST_App1) t1 t2 ==> t1' t2 value v1 t2 ==> t2'

---

(ST_App2) v1 t2 ==> v1 t2' ... plus the usual rules for booleans:

---

(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)

Examples

Example: ((\x:Bool->Bool. x) (\x:Bool. x)) ==>* (\x:Bool. x) i.e. (idBB idB) ==>* idB

Example: ((\x:Bool->Bool. x) ((\x:Bool->Bool. x) (\x:Bool. x))) ==>* (\x:Bool. x) i.e. (idBB (idBB idB)) ==>* idB.

Example: ((\x:Bool->Bool. x) (\x:Bool. if x then false else true)) true) ==>* false i.e. ((idBB notB) ttrue) ==>* tfalse.

Example: ((\x:Bool->Bool. x) ((\x:Bool. if x then false else true) true)) ==>* false i.e. (idBB (notB ttrue)) ==>* tfalse.

A more automatic proof

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.

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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 Gamma |- t \in T, where Gamma 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.

Informally, we'll write Gamma, x:T for "extend the partial function Gamma to also map x to T." Formally, we use the function extend to add a binding to a partial map.

Typing Relation

Gamma x = T

---

(T_Var) Gamma |- x \in T Gamma , x:T11 |- t12 \in T12

---

(T_Abs) Gamma |- \x:T11.t12 \in T11->T12 Gamma |- t1 \in T11->T12 Gamma |- t2 \in T11

---

(T_App) Gamma |- t1 t2 \in T12

---

(T_True) Gamma |- true \in Bool

---

(T_False) Gamma |- false \in Bool Gamma |- t1 \in Bool Gamma |- t2 \in T Gamma |- t3 \in T

---

(T_If) Gamma |- if t1 then t2 else t3 \in T We can read the three-place relation Gamma |- t \in T as: "to the term t we can assign the type T using as types for the free variables of t the ones specified in the context Gamma."

Examples

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

Another example: empty |- \x:A. \y:A->A. y (y x)) \in A -> (A->A) -> A.

Exercise: 2 stars, optional (typing_example_2_full)

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

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Exercise: 2 stars (typing_example_3)

Formally prove the following typing derivation holds: empty |- \x:Bool->B. \y:Bool->Bool. \z:Bool. y (x z) \in T.

☐ 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 \x:Bool. \y:Bool, x y -- i.e., ~ exists T, empty |- \x:Bool. \y:Bool, x y : T.

Exercise: 3 stars, optional (typing_nonexample_3)

Another nonexample: ~ (exists S, exists T, empty |- \x:S. x x : T).

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