StlcThe 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 arise from the mechanisms of variable binding and substitution. It will take some work to deal with these.

Set Warnings "-notation-overridden,-parsing".
From Coq Require Import Strings.String.
From PLF Require Import Maps.
From PLF Require Import Smallstep.

Overview

The STLC is built on some collection of base types: booleans, numbers, strings, etc. The exact choice of base types doesn't matter much -- the construction of the language and its theoretical properties work out the same no matter what we choose -- 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 boolean constants and conditionals, we add three things:
  • variables
  • function abstractions
  • application
This gives us the following collection of abstract syntax constructors (written out first in informal BNF notation -- we'll formalize it below).

       t ::= x (variable)
           | \x:T,t (abstraction)
           | t t (application)
           | true (constant true)
           | false (constant false)
           | if t then t else t (conditional)
The \ symbol in a function abstraction \x:T,t 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 t is its body. The annotation :T specifies the type of arguments that the function can be applied to.
If you've seen lambda-calculus notation before, you might be wondering why abstraction is written here as \x:T,t instead of the usual "\x:T.t". The reason is that some front ends for interacting with Coq use period to separate a file into "sentences" to be passed separately to the Coq top level.
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. (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.
    As in Coq, application associates to the left -- i.e., this expression is parsed as ((\x:Bool, \y:Bool, x) false) true.
  • \f:BoolBool, 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:BoolBool, 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.
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
          | TT
For example:
  • \x:Bool, false has type BoolBool
  • \x:Bool, x has type BoolBool
  • (\x:Bool, x) true has type Bool
  • \x:Bool, \y:Bool, x has type BoolBoolBool (i.e., Bool (BoolBool))
  • (\x:Bool, \y:Bool, x) false has type BoolBool
  • (\x:Bool, \y:Bool, x) false true has type Bool

Syntax

We next formalize the syntax of the STLC.

Module STLC.

Types


Inductive ty : Type :=
  | Ty_Bool : ty
  | Ty_Arrow : ty ty ty.

Terms


Inductive tm : Type :=
  | tm_var : string tm
  | tm_app : tm tm tm
  | tm_abs : string ty tm tm
  | tm_true : tm
  | tm_false : tm
  | tm_if : tm tm tm tm.

Declare Custom Entry stlc.
Notation "<{ e }>" := e (e custom stlc at level 99).
Notation "( x )" := x (in custom stlc, x at level 99).
Notation "x" := x (in custom stlc at level 0, x constr at level 0).
Notation "S -> T" := (Ty_Arrow S T) (in custom stlc at level 50, right associativity).
Notation "x y" := (tm_app x y) (in custom stlc at level 1, left associativity).
Notation "\ x : t , y" :=
  (tm_abs x t y) (in custom stlc at level 90, x at level 99,
                     t custom stlc at level 99,
                     y custom stlc at level 99,
                     left associativity).
Coercion tm_var : string >-> tm.

Notation "'Bool'" := Ty_Bool (in custom stlc at level 0).
Notation "'if' x 'then' y 'else' z" :=
  (tm_if x y z) (in custom stlc at level 89,
                    x custom stlc at level 99,
                    y custom stlc at level 99,
                    z custom stlc at level 99,
                    left associativity).
Notation "'true'" := true (at level 1).
Notation "'true'" := tm_true (in custom stlc at level 0).
Notation "'false'" := false (at level 1).
Notation "'false'" := tm_false (in custom stlc at level 0).
Some more notation magic to set up the concrete syntax, as we did in the Imp chapter...

Definition x : string := "x".
Definition y : string := "y".
Definition z : string := "z".
Hint Unfold x : core.
Hint Unfold y : core.
Hint Unfold z : core.
Note that an abstraction \x:T,t (formally, tm_abs 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.
Some examples...

Notation idB :=
  <{\x:Bool, x}>.

Notation idBB :=
  <{\x:BoolBool, x}>.

Notation idBBBB :=
  <{\x:((BoolBool)(BoolBool)), x}>.

Notation k := <{\x:Bool, \y:Bool, x}>.

Notation 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 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, t is a value only when t 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, t is always a value, no matter whether t is one or not -- in other words, we can say that reduction stops at abstractions.
Our usual way of evaluating expressions in Gallina makes the first choice -- for example,
         Compute (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.

Inductive value : tm Prop :=
  | v_abs : x T2 t1,
      value <{\x:T2, t1}>
  | v_true :
      value <{true}>
  | v_false :
      value <{false}>.

Hint Constructors value : core.
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 complete program is closed -- that is, it contains no free variables.
(Conversely, a term with free variables is often called an open term.)
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.

Substitution

Now we come to the heart of the STLC: the operation of substituting one term for a variable in another term. This operation is 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 s for x 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:=s]x = s
       [x:=s]y = y if xy
       [x:=s](\x:T, t) = \x:T, t
       [x:=s](\y:T, t) = \y:T, [x:=s]t if xy
       [x:=s](t1 t2) = ([x:=s]t1) ([x:=s]t2)
       [x:=s]true = true
       [x:=s]false = false
       [x:=s](if t1 then t2 else t3) =
              if [x:=s]t1 then [x:=s]t2 else [x:=s]t3
... and formally:

Reserved Notation "'[' x ':=' s ']' t" (in custom stlc at level 20, x constr).

Fixpoint subst (x : string) (s : tm) (t : tm) : tm :=
  match t with
  | tm_var y
      if eqb_string x y then s else t
  | <{\y:T, t1}>
      if eqb_string x y then t else <{\y:T, [x:=s] t1}>
  | <{t1 t2}>
      <{([x:=s] t1) ([x:=s] t2)}>
  | <{true}>
      <{true}>
  | <{false}>
      <{false}>
  | <{if t1 then t2 else t3}>
      <{if ([x:=s] t1) then ([x:=s] t2) else ([x:=s] t3)}>
  end

where "'[' x ':=' s ']' t" := (subst x s t) (in custom stlc).
Note on notations: You might be wondering why we need curly braces around the substitution notation in the above definition, and why do we need to redefine the substition notation in the stlc custom grammar. The reason is that reserved notations in definitions have to be defined in the general Coq grammar (and not a custom one like stlc). This restriction only applies to the subst definition, that is before the where ... part. From now on, using the substitution notation in the stlc custom grammar doesn't need any curly braces.
For example...
Check <{[x:=true] x}>.
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 include binders for all of the variables they mention), we can sidestep this extra complexity, but it must be dealt with when formalizing richer languages.
For example, using the definition of substitution above to substitute the open term s = \x:Bool, r, where r is a free reference to some global resource, for the variable z in the term t = \r:Bool, z, where r is a bound variable, we would get \r:Bool, \x:Bool, r, where the free reference to r in s has been "captured" by the binder at the beginning of t.
Why would this be bad? Because it violates the principle that the names of bound variables do not matter. For example, if we rename the bound variable in t, e.g., let t' = \w:Bool, z, then [x:=s]t' is \w:Bool, \x:Bool, r, which does not behave the same as [x:=s]t = \r:Bool, \x:Bool, r. That is, renaming a bound variable changes how t behaves under substitution.
See, for example, [Aydemir 2008] for further discussion of this issue.

Exercise: 3 stars, standard (substi_correct)

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 construectors; your job is to fill in the rest of the constructors and prove that the relation you've defined coincides with the function given above.

Inductive substi (s : tm) (x : string) : tm tm Prop :=
  | s_var1 :
      substi s x (tm_var x) s
  (* FILL IN HERE *)
.

Hint Constructors substi : core.

Theorem substi_correct : s x t t',
  <{ [x:=s]t }> = t' substi s x t t'.
Proof.
  (* FILL IN HERE *) Admitted.

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 (the function) until it becomes an abstraction; 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 abstraction. This last rule, written informally as
      (\x:T,t12) v2 --> [x:=v2]t12
is traditionally called beta-reduction.
value v2 (ST_AppAbs)  

(\x:T2,t1) v2 --> [x:=v2]t1
t1 --> t1' (ST_App1)  

t1 t2 --> t1' t2
value v1
t2 --> t2' (ST_App2)  

v1 t2 --> v1 t2'
... plus the usual rules for conditionals:
   (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)
Formally:

Reserved Notation "t '-->' t'" (at level 40).

Inductive step : tm tm Prop :=
  | ST_AppAbs : x T2 t1 v2,
         value v2
         <{(\x:T2, t1) v2}> --> <{ [x:=v2]t1 }>
  | ST_App1 : t1 t1' t2,
         t1 --> t1'
         <{t1 t2}> --> <{t1' t2}>
  | ST_App2 : v1 t2 t2',
         value v1
         t2 --> t2'
         <{v1 t2}> --> <{v1 t2'}>
  | ST_IfTrue : t1 t2,
      <{if true then t1 else t2}> --> t1
  | ST_IfFalse : t1 t2,
      <{if false then t1 else t2}> --> t2
  | ST_If : t1 t1' t2 t3,
      t1 --> t1'
      <{if t1 then t2 else t3}> --> <{if t1' then t2 else t3}>

where "t '-->' t'" := (step t t').

Hint Constructors step : core.

Notation multistep := (multi step).
Notation "t1 '-->*' t2" := (multistep t1 t2) (at level 40).

Examples

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

Lemma step_example1 :
  <{idBB idB}> -->* idB.
Proof.
  eapply multi_step.
    apply ST_AppAbs.
    apply v_abs.
  simpl.
  apply multi_refl. Qed.
Example:
      (\x:BoolBool, x) ((\x:BoolBool, x) (\x:Bool, x))
            -->* \x:Bool, x
i.e.,
      (idBB (idBB idB)) -->* idB.

Lemma step_example2 :
  <{idBB (idBB idB)}> -->* idB.
Proof.
  eapply multi_step.
    apply ST_App2. auto.
    apply ST_AppAbs. auto.
  eapply multi_step.
    apply ST_AppAbs. simpl. auto.
  simpl. apply multi_refl. Qed.
Example:
      (\x:BoolBool, x)
         (\x:Bool, if x then false else true)
         true
            -->* false
i.e.,
       (idBB notB) true -->* false.

Lemma step_example3 :
  <{idBB notB true}> -->* <{false}>.
Proof.
  eapply multi_step.
    apply ST_App1. apply ST_AppAbs. auto. simpl.
  eapply multi_step.
    apply ST_AppAbs. auto. simpl.
  eapply multi_step.
    apply ST_IfTrue. apply multi_refl. Qed.
Example:
      (\x:BoolBool, x)
         ((\x:Bool, if x then false else true) true)
            -->* false
i.e.,
      idBB (notB true) -->* false.
(Note that this term doesn't actually typecheck; even so, we can ask how it reduces.)

Lemma step_example4 :
  <{idBB (notB true)}> -->* <{false}>.
Proof.
  eapply multi_step.
    apply ST_App2. auto.
    apply ST_AppAbs. auto. simpl.
  eapply multi_step.
    apply ST_App2. auto.
    apply ST_IfTrue.
  eapply multi_step.
    apply ST_AppAbs. auto. simpl.
  apply multi_refl. Qed.
We can use the normalize tactic defined in the Smallstep chapter to simplify these proofs.

Lemma step_example1' :
  <{idBB idB}> -->* idB.
Proof. normalize. Qed.

Lemma step_example2' :
  <{idBB (idBB idB)}> -->* idB.
Proof. normalize. Qed.

Lemma step_example3' :
  <{idBB notB true}> -->* <{false}>.
Proof. normalize. Qed.

Lemma step_example4' :
  <{idBB (notB true)}> -->* <{false}>.
Proof. normalize. Qed.

Exercise: 2 stars, standard (step_example5)

Try to do this one both with and without normalize.

Lemma step_example5 :
       <{idBBBB idBB idB}>
  -->* idB.
Proof.
  (* FILL IN HERE *) Admitted.

Lemma step_example5_with_normalize :
       <{idBBBB idBB idB}>
  -->* idB.
Proof.
  (* FILL IN HERE *) Admitted.

Typing

Next we consider the typing relation of the STLC.

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.
Following the usual notation for partial maps, we write (X > T, Gamma) for "update the partial function Gamma so that it maps x to T."

Definition context := partial_map ty.

Typing Relation

Gamma x = T1 (T_Var)  

Gamma ⊢ x ∈ T1
x > T2 ; Gamma ⊢ t1 ∈ T1 (T_Abs)  

Gamma ⊢ \x:T2,t1 ∈ T2->T1
Gamma ⊢ t1 ∈ T2->T1
Gamma ⊢ t2 ∈ T2 (T_App)  

Gamma ⊢ t1 t2 ∈ T1
   (T_True)  

Gamma ⊢ true ∈ Bool
   (T_False)  

Gamma ⊢ false ∈ Bool
Gamma ⊢ t1 ∈ Bool    Gamma ⊢ t2 ∈ T1    Gamma ⊢ t3 ∈ T1 (T_If)  

Gamma ⊢ if t1 then t2 else t3 ∈ T1
We can read the three-place relation Gamma t \in T as: "under the assumptions in Gamma, the term t has the type T."

Reserved Notation "Gamma '⊢' t '∈' T" (at level 101,
                                          t custom stlc, T custom stlc at level 0).

Inductive has_type : context tm ty Prop :=
  | T_Var : Gamma x T1,
      Gamma x = Some T1
      Gamma x \in T1
  | T_Abs : Gamma x T1 T2 t1,
      x > T2 ; Gamma t1 \in T1
      Gamma \x:T2, t1 \in (T2 T1)
  | T_App : T1 T2 Gamma t1 t2,
      Gamma t1 \in (T2 T1)
      Gamma t2 \in T2
      Gamma t1 t2 \in T1
  | T_True : Gamma,
       Gamma true \in Bool
  | T_False : Gamma,
       Gamma false \in Bool
  | T_If : t1 t2 t3 T1 Gamma,
       Gamma t1 \in Bool
       Gamma t2 \in T1
       Gamma t3 \in T1
       Gamma if t1 then t2 else t3 \in T1

where "Gamma '⊢' t '∈' T" := (has_type Gamma t T).

Hint Constructors has_type : core.

Examples


Example typing_example_1 :
  empty \x:Bool, x \in (Bool Bool).
Proof.
  apply T_Abs. apply T_Var. reflexivity. Qed.
Note that, since we added the has_type constructors to the hints database, auto can actually solve this one immediately.

Example typing_example_1' :
  empty \x:Bool, x \in (Bool Bool).
Proof. auto. Qed.
More examples:
       empty ⊢ \x:A, \y:AA, y (y x)
             \in A → (AA) → A.

Example typing_example_2 :
  empty
    \x:Bool,
       \y:BoolBool,
          (y (y x)) \in
    (Bool (Bool Bool) Bool).
Proof.
  apply T_Abs.
  apply T_Abs.
  eapply T_App. apply T_Var. apply update_eq.
  eapply T_App. apply T_Var. apply update_eq.
  apply T_Var. apply update_neq. intros Contra. discriminate.
Qed.

Exercise: 2 stars, standard, optional (typing_example_2_full)

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

Example typing_example_2_full :
  empty
    \x:Bool,
       \y:BoolBool,
          (y (y x)) \in
    (Bool (Bool Bool) Bool).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard (typing_example_3)

Formally prove the following typing derivation holds:
       empty ⊢ \x:BoolB, \y:BoolBool, \z:Bool,
                   y (x z)
             \in T.

Example typing_example_3 :
   T,
    empty
      \x:BoolBool,
         \y:BoolBool,
            \z:Bool,
               (y (x z)) \in
      T.
Proof.
  (* FILL IN HERE *) Admitted.
We can also show that some terms are not typable. For example, let's check that there is no typing derivation assigning a type to the term \x:Bool, \y:Bool, x y -- i.e.,
    ¬ T,
        empty ⊢ \x:Bool, \y:Bool, x y \in T.

Example typing_nonexample_1 :
  ¬ T,
      empty
        \x:Bool,
            \y:Bool,
               (x y) \in
        T.
Proof.
  intros Hc. destruct Hc as [T Hc].
  (* The clear tactic is useful here for tidying away bits of
     the context that we're not going to need again. *)

  inversion Hc; subst; clear Hc.
  inversion H4; subst; clear H4.
  inversion H5; subst; clear H5 H4.
  inversion H2; subst; clear H2.
  discriminate H1.
Qed.

Exercise: 3 stars, standard, optional (typing_nonexample_3)

Another nonexample:
    ¬( S T,
          empty ⊢ \x:S, x x \in T).

Example typing_nonexample_3 :
  ¬ ( S T,
        empty
          \x:S, x x \in T).
Proof.
  (* FILL IN HERE *) Admitted.

End STLC.

(* 2020-09-09 21:08 *)