I'm talking about first-order existential types, as summarized here, with pack as the introduction form that bundles up a type with an expression of that type, and open as the elimination form that unpacks the type and its representative expression.

This seems just as fundamental as parametric polymorphism (universal types), so I'm surprised that this construct isn't implemented in e.g. Haskell or OCaml. Does anyone know the reasons for this?

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    $\begingroup$ More details are needed. When one says existential type do they not mean a witness and a proof that the witness satisfies some property? If so, what is a "proof" in imperative/OO languages? $\endgroup$ Nov 8, 2016 at 21:12
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    $\begingroup$ @MusaAl-hassy: No, they do not mean that, see stackoverflow.com/questions/292274/what-is-an-existential-type $\endgroup$ Nov 8, 2016 at 21:48
  • $\begingroup$ Learning things is a delight! Now I've more enjoyable articles to read; Thanks Andrej! $\endgroup$ Nov 9, 2016 at 14:28
  • $\begingroup$ Here is a bit of explanation to the less sophisticated readers who may want to understand the comments here. Constructively, we convince others that $\exists x.\varphi(x)$ holds if we can show an example (i.e., witness) $a$ and present an argument that $\varphi(a)$ holds. In some settings, we can consider types as logical statements, and terms as proofs to those statements. That is the background of Al-hassy's question. One possible choice of further reading may be Paul Taylor's Practical Foundations of Mathematics, Section 2.4. $\endgroup$
    – Yuning
    Feb 5, 2021 at 14:09
  • $\begingroup$ Nevertheless, for the moment very few real-life programming languages possess such immediate correspondence to logic. Typed functional languages are usually good approximations, while imperative/OO languages are probably too far apart. For that reason witness-proof is technically not a way to think of `existential types' in imperative languages. However, forming and using existential types there involve a type $a$ and a value of type $\varphi(a)$ in a similar way. So to a programmer trying to express abstraction in their data and code, the mechanism is similar. $\endgroup$
    – Yuning
    Feb 5, 2021 at 14:19

1 Answer 1


OCaml's idea of modules roughly correspond with existential types ;)

Let's set up some rough semantics in order to frame this correspondence properly. First, I'll give the intro/elimination rules for existential types (I'll only use a variable $\to$ type context $\Gamma$ for the sake of simplicity here): $$ \frac{\Gamma \vdash e\{\alpha\mapsto \tau\} : \sigma\{\alpha\mapsto \tau\}}{\Gamma \vdash [\tau, e] : \exists \alpha. \sigma} ({\rm Pack}) ~~~~~~~ \frac{\Gamma \vdash e : \exists \alpha.\sigma ~~~ \Gamma, x:\sigma \vdash e' : \tau'}{\Gamma \vdash {\rm let} ~ [\alpha, x] = e ~ {\rm in} ~ e' : \tau'}({\rm Unpack}) $$ where $e\{\alpha \mapsto \tau\}$ substitutes all occurrences of the type variable $\alpha$ with the concrete type $\tau$. In practice, we will also decorate the packed form with its existential type in order to disambiguate the semantics.

Operationally, we have the context reductions: $$ \frac{e \Downarrow v}{[\tau, e] \Downarrow [\tau, v]} ~~~~~~ \frac{e \Downarrow v}{{\rm let} ~ [\alpha, x] = e ~ {\rm in} ~ e' \Downarrow {\rm let} ~ [\alpha, x] = v ~ {\rm in} ~ e'} $$ which corresponds to the reduction strategies that reduces the expressions in the existential packages and the expressions to be unpacked.

Finally, we have the extensional $\eta$-reduction rule: $$ \frac{}{{\rm let} ~ [\alpha, x] = [\tau, v]_{\exists \alpha. \sigma} ~ {\rm in} ~ e \to e \{\alpha \mapsto \tau, x \mapsto v\}} $$

So for example, the program \begin{align*} {\rm let} ~ [\alpha, x] &= [{\rm int}, (0, {\rm succ})]_{\exists \alpha. \alpha \times (\alpha \to \alpha)} ~ {\rm in} ~ ({\rm snd} ~ x : \alpha \to \alpha)({\rm fst} ~ x : \alpha)\\ &\to^{\eta} ({\rm snd} ~ (0, {\rm succ}) : {\rm int} \to {\rm int})({\rm fst} ~ (0, {\rm succ}) : {\rm int}) \\ &\to^* {\rm succ} ~ 0 \\ &\to^\beta 1 \end{align*}

This exact idea is present in OCaml's concept of modules, save that the packed type must be a record (but which can be made to be isomorphic to any other OCaml type). For example, the creation of an existential type is done through the module type keyword:

module type Bundle =
  type α
  val data : α * (α -> α)

This signature corresponds to the existential type $\exists \alpha. \alpha \times (\alpha \to \alpha)$. We can witness an inhabitant of this type by instantiating it with the struct keyword:

module BundleInt : Bundle = struct
  type α = int
  val data = (0, (+) 1)

which corresponds to the pack/intro $[{\rm int}, (0, (+) 1)]_{\exists \alpha. \alpha \times (\alpha \to \alpha)}$. Unfortunately, there is no explicit ${\rm let} \dots$ elimination form, but something like open BundleInt is equivalent to let [α, data] = BundleInt in ... where α and data are specified in the creation of the module type Bundle.

So while module types aren't exactly the same as your traditional existential types (the variables to be bound are explicitly pushed into the intro-form rather than in the elimination-form), their expressiveness are the same.

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    $\begingroup$ For some reason I don't remember seeing this answer, but thanks nonetheless! There actually is an let open construct in OCaml which seems to have nearly the same semantics as unpack. That said, there is one point I disagree with; you say that they're equally expressive, but with existential types, you can write a function parameterized by an existential type, whereas in OCaml you can't (you have to write functors parameterized by modules). $\endgroup$
    – gardenhead
    Jan 27, 2017 at 5:33

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