[TYPES] Subtyping, extensional equality, and contravariance

Tim Sweeney Tim.Sweeney at epicgames.com
Mon Jul 25 22:40:19 EDT 2005

This is a query about a new direction in type systems supporting
subtyping, extensional equality, and contravariance.

Consider a type system with types "int" (the integers) and "nat" (the
natural numbers) with nat<:int. Now define:

	abs :: int -> int
	abs x = if x >= 0 then x else -x

	id :: int -> int
	id x = x

Traditionally, one would say that abs!=int (at type int->int), and
abs=id (at type nat->nat). In other words, extensional equality is only
meaningful "at a type t", and the extensional equality of values may
differ depending on the types at which they are interpretted.

But I am interested in a type system in which equality represents the
same relation at all types, in other words: a=b (at type t) implies that
a=b (at type u) for all types u containing a and b.  Thus we can talk of
a=b without referring to an interpretation type t.  The effect of this
is to make all values non-polymorphic, that is, their interpretation is
independent of the context in which they are interpretted.

To satisfy this goal, I found it necessary to associate with every
function value a unique "domain type" which is extractable and is
carried around with the function value:

	dom :: forall t,u. t -> u -> type
	dom f = t

Then, two functions f and g are equal iff dom(f)=dom(g) and for all
x:dom(f), f(x)=g(x).  In the examples above, dom(abs)=dom(id)=int, and

The "invariant function space" type (written t+>u for now) then means
the type of functions f in which dom(f)=t and for all x:t, f(x):u.  Thus
we have abs:int+>int, but not abs:nat+>nat, because dom(abs)=int.

The "contravariant function space" type t->u then means (intuitively)
the union, for all supertypes types t' of t and all types u', of all
functions f:t'+>u' in which for all x:t, f(x) belongs to type u'. In
other words, all functions whose domain includes t, and whose image
under t belongs to u. This is interesting because we then have
abs:int->int and abs:nat->nat, but not abs:int->nat.

Given an empty type "empty" and any type t, the type "empty->t" is then
the type of all functions; name the later "function".

This approach appears to enable implicit parametric polymorphism without
the existing of any (implicit or explicit "forall") quantifiers. For
example, in Haskell one would write:

	map :: (t -> u) -> [t] -> [u]
	map f xs = [f x | x <- xs]
	as = map abs [3,-5,7]

If we took the dependent-typed language Cayenne and removed parametric
polymorphism, to write this we would have to specify all types

	map (t::type) (u::type) (f::t->u) (xs::[t])
	as = map int int abs [3,-5,7]

But with the type system above, where every function value carries its
exact domain type along with it, we can write the following:

	map (f::function) (xs::[dom(f)]) = [f x | x <- xs]
	as = map abs [3,-5,7]

Note that the above "map abs" application is as concise as in Haskell,
but requires no forall quantifiers (neither explicit nor implicit), as
the function's domain type is extracted from the function's value.

We can also define a universal type "top" containing all possible
values.  And then a Haskell function like:

	id :: t -> t
	id x = x

Can be translated without quantifiers as:

	id (x::top) = x

Has this approach been taken before?  I would appreciate any references
to work along these lines, especially results on parametricity.

Tim Sweeney

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