skaller wrote:
> So actually, this F is distinct from Foo, even though all the
> members are the same. In particular, if Foo has an abstract type t,
> is F.t the same type as Foo.t?

Yes, because the signature inferred for F keeps a sharing constraint
t = Foo.t. But if G is some functor with an abstract type s in the
result, then G(F).s and G(Foo).s are not equal (because there is no
sharing constraint between modules, such as F = Foo).

-- Alain