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Software Technology: Simply-typed lambda calculus


This material is taken verbatim from Software Foundations V4.0

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Assumed Knowledge: * TODO
Learning Outcomes: * TODO

Require Import Maps.
Require Import Smallstep.
Require Import Types.

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


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:T1.t2                abstraction
           | t1 t2                   application
           | true                    constant true
           | false                   constant false
           | if t1 then t2 else t3   conditional
The \ symbol 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 t2 is its body. The annotation :T1 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. (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
          | T1  T2
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


We begin by formalizing the syntax of the STLC.

Module STLC.


Inductive ty : Type :=
  | TBool : ty
  | TArrow : ty ty ty.


Inductive tm : Type :=
  | tvar : id tm
  | tapp : tm tm tm
  | tabs : id ty tm tm
  | ttrue : tm
  | tfalse : tm
  | tif : tm tm tm tm.

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.
Some examples...

Definition x := (Id 0).
Definition y := (Id 1).
Definition z := (Id 2).
Hint Unfold x.
Hint Unfold y.
Hint Unfold z.

idB = \x:Bool. x

Notation idB :=
  (tabs x TBool (tvar x)).

idBB = \x:BoolBool. x

Notation idBB :=
  (tabs x (TArrow TBool TBool) (tvar x)).

idBBBB = \x:(BoolBool) (BoolBool). x

Notation idBBBB :=
  (tabs x (TArrow (TArrow TBool TBool)
                      (TArrow TBool TBool))
    (tvar x)).

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

Notation k := (tabs x TBool (tabs y TBool (tvar x))).

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

Notation notB := (tabs x TBool (tif (tvar x) tfalse ttrue)).

(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.


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 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 T t,
      value (tabs x T t)
  | v_true :
      value ttrue
  | v_false :
      value tfalse.

Hint Constructors value.

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.
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.


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 xfalse
       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:=s]x               = s
       [x:=s]y               = y                      if x ≠ y
       [x:=s](\x:T11. t12)   = \x:T11. t12
       [x:=s](\y:T11. t12)   = \y:T11. [x:=s]t12      if x ≠ y
       [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" (at level 20).

Fixpoint subst (x:id) (s:tm) (t:tm) : tm :=
  match t with
  | tvar x'
      if beq_id x x' then s else t
  | tabs x' T t1
      tabs x' T (if beq_id x x' then t1 else ([x:=s] t1))
  | tapp t1 t2
      tapp ([x:=s] t1) ([x:=s] t2)
  | ttrue
  | tfalse
  | tif t1 t2 t3
      tif ([x:=s] t1) ([x:=s] t2) ([x:=s] t3)

where "'[' x ':=' s ']' t" := (subst x s t).

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 avoid 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 and prove that the relation you've defined coincides with the function given above.

Inductive substi (s:tm) (x:id) : tm tm Prop :=
  | s_var1 :
      substi s x (tvar x) s
  (* FILL IN HERE *)

Hint Constructors substi.

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


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.t12v2  [x:=v2]t12
is traditionally called "beta-reduction".
value v2 (ST_AppAbs)  

(\x:T.t12) v2  [x:=v2]t12
t1  t1' (ST_App1)  

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

v1 t2  v1 t2'
... plus the usual rules for booleans:

(if true then t1 else t2 t1

(if false then t1 else t2 t2
t1  t1' (ST_If)  

(if t1 then t2 else t3 (if t1' then t2 else t3)

Reserved Notation "t1 '' t2" (at level 40).

Inductive step : tm tm Prop :=
  | ST_AppAbs : x T t12 v2,
         value v2
         (tapp (tabs x T t12) v2) [x:=v2]t12
  | ST_App1 : t1 t1' t2,
         t1 t1'
         tapp t1 t2 tapp t1' t2
  | ST_App2 : v1 t2 t2',
         value v1
         t2 t2'
         tapp v1 t2 tapp v1 t2'
  | ST_IfTrue : t1 t2,
      (tif ttrue t1 t2) t1
  | ST_IfFalse : t1 t2,
      (tif tfalse t1 t2) t2
  | ST_If : t1 t1' t2 t3,
      t1 t1'
      (tif t1 t2 t3) (tif t1' t2 t3)

where "t1 '' t2" := (step t1 t2).

Hint Constructors step.

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


      ((\x:BoolBool. x) (\x:Bool. x)) ⇒* (\x:Bool. x)
      (idBB idB⇒* idB

Lemma step_example1 :
  (tapp idBB idB) ⇒* idB.
  eapply multi_step.
    apply ST_AppAbs.
    apply v_abs.
  apply multi_refl. Qed.

      ((\x:BoolBool. x) ((\x:BoolBool. x) (\x:Bool. x)))
            ⇒* (\x:Bool. x)
      (idBB (idBB idB)) ⇒* idB.

Lemma step_example2 :
  (tapp idBB (tapp idBB idB)) ⇒* idB.
  eapply multi_step.
    apply ST_App2. auto.
    apply ST_AppAbs. auto.
  eapply multi_step.
    apply ST_AppAbs. simpl. auto.
  simpl. apply multi_refl. Qed.

      ((\x:BoolBool. x) (\x:Bool. if x then false
                                    else true)) true)
            ⇒* false
       ((idBB notBttrue⇒* tfalse.

Lemma step_example3 :
  tapp (tapp idBB notB) ttrue ⇒* tfalse.
  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.

((\x:Bool  Bool. x) ((\x:Bool. if x then false
                               else truetrue))
      ⇒* false
  (idBB (notB ttrue)) ⇒* tfalse.

Lemma step_example4 :
  tapp idBB (tapp notB ttrue) ⇒* tfalse.
  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 Types chapter to simplify these proofs.

Lemma step_example1' :
  (tapp idBB idB) ⇒* idB.
Proof. normalize. Qed.

Lemma step_example2' :
  (tapp idBB (tapp idBB idB)) ⇒* idB.
Proof. normalize. Qed.

Lemma step_example3' :
  tapp (tapp idBB notB) ttrue ⇒* tfalse.
Proof. normalize. Qed.

Lemma step_example4' :
  tapp idBB (tapp notB ttrue) ⇒* tfalse.
Proof. normalize. Qed.

Exercise: 2 stars (step_example3)

Try to do this one both with and without normalize.

Lemma step_example5 :
       (tapp (tapp idBBBB idBB) idB)
  ⇒* idB.
  (* FILL IN HERE *) Admitted.

Lemma step_example5_with_normalize :
       (tapp (tapp idBBBB idBB) idB)
  ⇒* idB.
  (* FILL IN HERE *) Admitted.


Next we consider the typing relation of the STLC.


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

Definition context := partial_map ty.

Typing Relation

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

Γ  true ∈ Bool

Γ  false ∈ Bool
Γ  t1 ∈ Bool    Γ  t2 ∈ T    Γ  t3 ∈ T (T_If)  

Γ  if t1 then t2 else t3 ∈ T
We can read the three-place relation Γ t 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 Γ."

Reserved Notation "Gamma '' t '∈' T" (at level 40).

Inductive has_type : context tm ty Prop :=
  | T_Var : Γ x T,
      Γ x = Some T
      Γ tvar xT
  | T_Abs : Γ x T11 T12 t12,
      update Γ x T11 t12T12
      Γ tabs x T11 t12TArrow T11 T12
  | T_App : T11 T12 Γ t1 t2,
      Γ t1TArrow T11 T12
      Γ t2T11
      Γ tapp t1 t2T12
  | T_True : Γ,
       Γ ttrueTBool
  | T_False : Γ,
       Γ tfalseTBool
  | T_If : t1 t2 t3 T Γ,
       Γ t1TBool
       Γ t2T
       Γ t3T
       Γ tif t1 t2 t3T

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

Hint Constructors has_type.


Example typing_example_1 :
  empty tabs x TBool (tvar x) ∈ TArrow TBool TBool.
  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 tabs x TBool (tvar x) ∈ TArrow TBool TBool.
Proof. auto. Qed.

Another example:
       empty  \x:A. λy:AA. y (y x))
             ∈ A  (AA A.

Example typing_example_2 :
    (tabs x TBool
       (tabs y (TArrow TBool TBool)
          (tapp (tvar y) (tapp (tvar y) (tvar x))))) ∈
    (TArrow TBool (TArrow (TArrow TBool TBool) TBool)).
Proof with auto using update_eq.
  apply T_Abs.
  apply T_Abs.
  eapply T_App. apply T_Var...
  eapply T_App. apply T_Var...
  apply T_Var...

Exercise: 2 stars, optional (typing_example_2_full)

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

Example typing_example_2_full :
    (tabs x TBool
       (tabs y (TArrow TBool TBool)
          (tapp (tvar y) (tapp (tvar y) (tvar x))))) ∈
    (TArrow TBool (TArrow (TArrow TBool TBool) TBool)).
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars (typing_example_3)

Formally prove the following typing derivation holds:
       empty  \x:BoolB. λy:BoolBool. λz:Bool.
                   y (x z)
             ∈ T.

Example typing_example_3 :
      (tabs x (TArrow TBool TBool)
         (tabs y (TArrow TBool TBool)
            (tabs z TBool
               (tapp (tvar y) (tapp (tvar x) (tvar z)))))) ∈
Proof with auto.
  (* FILL IN HERE *) Admitted.
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.,
    ¬ T,
        empty  \x:Bool. λy:Boolx y : T.

Example typing_nonexample_1 :
  ¬ T,
        (tabs x TBool
            (tabs y TBool
               (tapp (tvar x) (tvar y)))) ∈
  intros Hc. inversion Hc.
  (* The clear tactic is useful here for tidying away bits of
     the context that we're not going to need again. *)

  inversion H. subst. clear H.
  inversion H5. subst. clear H5.
  inversion H4. subst. clear H4.
  inversion H2. subst. clear H2.
  inversion H5. subst. clear H5.
  inversion H1. Qed.

Exercise: 3 stars, optional (typing_nonexample_3)

Another nonexample:
    ¬ (ST,
          empty  \x:S. x x ∈ T).

Example typing_nonexample_3 :
  ¬ (S, T,
          (tabs x S
             (tapp (tvar x) (tvar x))) ∈
  (* FILL IN HERE *) Admitted.