I'm working on a small lambda calculus compiler that has a working Hindley-Milner type inference system and now also supports recursive let's (not in the linked code), which I understand should be enough to make it Turing complete.

The problem now is I have no idea how to make it support lists, or whether it already does support them and I just need to find a way to encode them. I'd like to be able to define them without having to add new rules to the type system.

The easiest way I can think of a list of x is as something that is either null (or the empty list), or a pair that contains both an x and a list of x. But to do this I need to be able to define pairs and or's, which I believe are the product and the sum types.

Seems that I can define pairs this way:

pair = λabf.fab
first = λp.p(λab.a)
second = λp.p(λab.b)

Since pair would have the type a -> (b -> ((a -> (b -> x)) -> x)), after passing, say, an int and a string, it'd yield something with type (int -> (string -> x)) -> x, which would be the representation of a pair of int and string. What bothers me here is that if that represents a pair, why is that not logically equivalent to, nor implies the proposition int and string?. However, is equivalent to (((int and string) -> x) -> x), as if I could only have product types as parameters to functions. This answer seem to address this problem, but I have no idea what the symbols he uses mean. Also, if this does not really encode a product type, is there anything I can do with product types I couldn't do with my definition of pairs above (considering I can also define n-tuples the same way)? If not, wouldn't this contradict the fact that you cannot express (AFAIK) conjunction using only implication?

Also, how about the sum type? Can I somehow encode it using only the function type? If so, would this be enough to define lists? Or else, is there any other way to define lists without having to extend my type system? And if not, what changes would I need to make if I want to keep it as simple as possible?

Please keep in mind that I'm a computer programmer but not a computer scientist nor a mathematician and pretty bad at reading math notation.

Edit: I'm not sure what's the technical name of what I have implemented so far, but all I have is basically the code I've linked above, which is a constraint generation algorithm that uses the rules for applications, abstractions and variables taken from the Hinley-Milner algorithm and then a unification algorithm that gets the principal type. For instance, the expression \a.a will yield the type a -> a, and the expression \a.(a a) will throw an occurs check error. On top of this, there is not exactly a let rule but a function that seems to have the same effect that lets you define recursive global functions like this pseudo-code:

GetTypeOfGlobalFunction(term, globalScope, nameOfFunction)
    // Here 'globalScope' contains a list of name-value pair where every value is of class 'ClosedType', 
    // meaning their type will be cloned before unified in the unification algorithm so that they can be used polymorphically 
    tempType = new TypeVariable() // Assign a dummy type to `tempType`, say, type 'x'.
    // The next line creates an scope with everything in 'globalScope' plus the 'nameOfFunction = tempType' name-value pair
    tempScope = new Scope(globalScope, nameOfFunction, tempType) 
    type = TypeOfTerm(term, tempScope) // Calculate the type of the term 
    Unify(tempType, type)
    return type
    // After returning, the code outside will create a 'ClosedType' using the returned type and add it to the global scope.

The code basically gets the type of the term as usual, but before unifying, it adds the name of the function being defined with a dummy type into the type scope so that it can be used from within itself recursively.

Edit 2: I just realized that I'd also need recursive types, which I don't have, to define a list like I want.

  • $\begingroup$ Can you be a bit more specific about what exactly you have implemented? Have you implemented the simply typed lambda calculus (with recursive definitions) and gave it parametric polymorphisms Hindley-Milner style? Or have you implemented the second-order polymorphic lambda calculus? $\endgroup$ Jan 30 '14 at 9:21
  • $\begingroup$ Maybe I can ask in an easier way: if I take OCaml or SML and restrict it to pure lambda terms and recursive definitions, is that what you're talking about? $\endgroup$ Jan 30 '14 at 9:22
  • $\begingroup$ @AndrejBauer: I've edited the question. I'm not sure about OCaml and SML, but I'm pretty sure if you take Haskell and restrict it to lambda terms and top level recursive lets (such as let func = \x -> (func x)) you get what I have. $\endgroup$
    – Juan
    Jan 30 '14 at 16:36
  • 1
    $\begingroup$ To maybe improve your question, checkout this meta post. $\endgroup$
    – Juho
    Jan 30 '14 at 21:11


This encoding is the Church encoding of pairs. Similar techniques can encode booleans, integers, lists, and other data structures.

Under the context x:a; y:b, the term pair x y has the type (a -> b -> t) -> t. The logical interpretation of this type is the following formula (I use standard mathematical notations: $\to$ is implication, $\vee$ is or, $\wedge$ is and, $\neg$ is negation; $\iff$ is equivalence of formulas): $$ \begin{align} (a \to b \to t) \to t & \quad\iff\quad \neg (\neg a \vee \neg b \vee t) \vee t \\ & \quad\iff\quad (a \wedge b \wedge \neg t) \vee t \\ & \quad\iff\quad (a \wedge b) \vee t \\ \end{align} $$ Why “a and b or t”? Intuitively, because pair is a function that takes a function as an argument and applies it to the pair. There are two ways this can go: the argument function can make use of the pair, or it can produce a value of type t without using the pair at all.

pair is a constructor for the pair type and first and second are destructors. (These are the same words used in object-oriented programming; here the words have a meaning that is related to the logical interpretation of the types and terms that I won't go into here.) Intuitively, the destructors let you access what is in the object and the constructors pave the way for the destructor by taking as an argument a function that they apply to the parts of the object. This principle can be applied to other types.


The Church encoding of a discriminated union is essentially dual to the Church encoding of a pair. Where a pair has two parts that must be put together and you can choose to extract one or the other, you can choose to build the union in either of two ways and when you use it you need to allow for both ways. Thus there are two constructors, and there is a single destructor which takes two arguments.

let case w = λf. λg. w f g           case : ((a->t) -> (b->t) -> t) -> (a->t) -> (b->t) -> t
  (* or simply let case w = w *)
let left x = λf. λg. f x             left : a -> ((a->t) -> (b->t) -> t)
let right y = λf. λg. g x            right : b -> ((a->t) -> (b->t) -> t)

Let me abbreviate the type (a->t) -> (b->t) -> t as SUM(a,b)(t). Then the types of the destructors and constructors are:

case : SUM(a,b)(t) -> (a->t) -> (b->t) -> t
left : a -> SUM(a,b)(t)
right : b -> SUM(a,b)(t)


case (left x) f g → f x
case (rightt y) f g → g y


For a list, apply again the same principle. A list whose elements have the type a can be built in two ways: it can be an empty list, or it can be an element (the head) plus a list (the tail). Compared with pairs, there's a little twist when it comes to the destructors: you can't have two separate destructors head and tail because they would only work on non-empty lists. You need a single destructor, with two arguments, one of which is a 0-argument function (i.e. a value) for the nil case, and the other a 2-argument function for the cons case. Functions like is_empty, head and tail can be derived from that. Like in the case of sums, the list is directly its own destructor function.

let nil = λn. λc. n
let cons h t = λn. λc. c h t
let is_empty l = l true (λh. λt. false) 
let head l default = l default (λh. λt. h)
let tail l default = l default (λh. λt. t)

Each of these functions is polymorphic. If you work the types of these functions, you'll note than cons is not uniform: the type of the result is not the same as the type of the argument. The type of the result is a variable — it is more general than the argument. If you chain cons calls, the successive calls to build a list instantiate cons at different types. This is crucial to make lists work in the absence of recursive types. You can make heterogeneous lists that way. In fact, the types that you can express are not “list of $T$”, but “list whose first elements are of types $T_1,\ldots,T_n$”.

As you surmise, if you want to define a type that only contains homogeneous lists, you need recursive types. Why? Let's look at the type of a list. A list is encoded as a function which takes two arguments: the value to return on empty lists, and the function to compute the value to return on a cons cell. Let a be the element type, b be the type of the list, and c be the type returned by the destructor. The type of a list is

a -> (a -> b -> c) -> c

Making the list homogeneous is saying that if it's a cons cell, the tail has to have the same type as the whole, i.e. it adds the constraint

a -> (a -> b -> c) -> c = b

The Hindley-Milner type system can be extended with such recursive types, and in fact practical programming languages do that. Practical programming languages tend to disallow such “naked” equations and require a data constructor, but this is not an intrinsic requirement of the underlying theory. Requiring a data constructor simplifies type inference, and in practice tends to avoids accepting functions that are actually buggy but happen to be typable with some unintended constraint that causes a difficult-to-understand type error where the function is used. This is why, for example, OCaml accepts unguarded recursive types only with the non-default -rectypes compiler option. Here are the definitions above in OCaml syntax, together with a type definition for homogeneous lists using the notation for aliased recursive types: type_expression as 'a means that the type type_expression is unified with the variable 'a.

# let nil = fun n c -> n;;
val nil : 'a -> 'b -> 'a = <fun>
# let cons h t = fun n c -> c h t;;
val cons : 'a -> 'b -> 'c -> ('a -> 'b -> 'd) -> 'd = <fun>
# let is_empty l = l true (fun h t -> false);;
val is_empty : (bool -> ('a -> 'b -> bool) -> 'c) -> 'c = <fun>
# let head l default = l default (fun h t -> h);;
val head : ('a -> ('b -> 'c -> 'b) -> 'd) -> 'a -> 'd = <fun>
# let tail l default = l default (fun h t -> t);;
val tail : ('a -> ('b -> 'c -> 'c) -> 'd) -> 'a -> 'd = <fun>
# type ('a, 'b, 'c) ulist = 'c -> ('a -> 'b -> 'c) -> 'c;;
type ('a, 'b, 'c) ulist = 'c -> ('a -> 'b -> 'c) -> 'c
# is_empty (cons 1 nil);;
- : bool = false
# head (cons 1 nil) 0;;
- : int = 1
# head (tail (cons 1 (cons 2.0 nil)) nil) 0.;;
- : float = 2.

(* -rectypes is required for what follows *)
# type ('a, 'b, 'c) rlist = 'c -> ('a -> 'b -> 'c) -> 'c as 'b;;
type ('a, 'b, 'c) rlist = 'b constraint 'b = 'c -> ('a -> 'b -> 'c) -> 'c
# let rcons = (cons : 'a -> ('a, 'b, 'c) rlist -> ('a, 'b, 'c) rlist);;
val rcons :
  'a ->
  ('a, 'c -> ('a -> 'b -> 'c) -> 'c as 'b, 'c) rlist -> ('a, 'b, 'c) rlist =
# head (rcons 1 (rcons 2 nil)) 0;;
- : int = 1
# tail (rcons 1 (rcons 2 nil)) nil;;
- : 'a -> (int -> 'a -> 'a) -> 'a as 'a = <fun>
# rcons 1 (rcons 2.0 nil);;
Error: This expression has type
         (float, 'b -> (float -> 'a -> 'b) -> 'b as 'a, 'b) rlist = 'a
       but an expression was expected of type
         (int, 'b -> (int -> 'c -> 'b) -> 'b as 'c, 'b) rlist = 'c


Looking at this a bit more generally, what is the function that represents the data structure?

  • For a natural number: $n$ is represented as the function which iterates its argument $n$ times.
  • For a pair: $(x,y)$ is represented as a function which applies its argument to $x$ and $y$.
  • For a sum: $\mathrm{in}_i(x)$ is represented as a function which applies its $i$th argument to $x$.
  • For a list: $[x_1,\ldots,x_n]$ is represented as a function that takes two arguments, the value to return for the empty list and the function to apply to cons cells.

In general terms, the data structure is represented as its fold function. This is a general concept for data structures: a fold function is a higher-order function that traverses the data structure. There is a technical sense in which fold is universal: all “generic” data structure traversals can be expressed in terms of fold. That the data structure can be represented as its fold function shows this: all you need to know about a data structure is how to traverse it, the rest is an implementation detail.

  • $\begingroup$ You mention “Church encoding” of integers, pairs, sums, but for lists you give the Scott encoding. I think it might be a bit confusing for those who are not familiar with encodings of inductive types. $\endgroup$ Jan 31 '14 at 14:19
  • $\begingroup$ So basically, my pair type is not really a product type as a function with this type could just return t and ignore the argument that is supposed to take a and b (which is exactly what (a and b) or t is saying). And seems that I'd have the same kind of trouble with sums. And also, without recursive types I won't have homogeneous list. So in a few words, are you saying I should add sum, product and recursive type rules to get homogeneous lists? $\endgroup$
    – Juan
    Jan 31 '14 at 16:04
  • $\begingroup$ Did you mean case (right y) f g → g y at the end of your Sums section? $\endgroup$
    – Juan
    Feb 3 '14 at 6:57
  • $\begingroup$ @StéphaneGimenez I hadn't realized. I'm not used to working on these encodings in a typed world. Can you give a reference for Church encoding vs Scott encoding? $\endgroup$ Feb 3 '14 at 12:03
  • $\begingroup$ @JuanLuisSoldi You've probably heard that “there is no problem that cannot be solved with an extra level of indirection”. Church encodings encode data structures as functions by adding a level of function call: a data structure becomes a second-order function that you apply to the function to act on the parts. If you want a homogeneous list type, you're going to have to deal with the fact that the type of the tail is the same as the type of the whole list. I think this has to involve a form of type recursion. $\endgroup$ Feb 3 '14 at 12:06

You can represent sum types as product types with tags and values. In this case, we can cheat a bit and use one tag to represent null or not, having the second tag representing the head/tail pair.

We define booleans in the usual way:

true = λi.λe.i
false = λi.λe.e
if = λcond.λthen.λelse.(cond then else)

A list is then a pair with the first element as a boolean, and the second element as a head/tail pair. Some basic list functions:

isNull = λl.(first l)
null = pair false false     --The second element doesn't matter in this case
cons = λh.λt.(pair true (pair h t ))
head = λl.(fst (snd l))   --This is a partial function
tail = λl.(snd (snd l))   --This is a partial function  

map = λf.λl.(if (isNull l)
                 (cons (f (head l)) (map f (tail l) ) ) 
  • $\begingroup$ But this wouldn't give me homogeneous list, is this correct? $\endgroup$
    – Juan
    Jan 31 '14 at 16:06

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