Yes, this still occurs in modern type theory; in particular, you'll find it in the calculus of constructions employed by the Coq language.
Consider the type called Prop
, whose inhabitants are logical propositions (which are in turn inhabited by proofs). The type Prop
does not belong to Prop
-- this means that Prop
exhibits stratification:
Check Prop.
Prop
: Type
However, note that (forall a:Prop, a)
does have type Prop
. So although Prop
does not belong to Prop
, things which quantify over all of Prop
may still belong to Prop
. So we can be more specific and say that Prop
exhibits unramified stratification.
Check (forall a:Prop, a).
forall a : Prop, a
: Prop
By contrast, consider Set
, whose inhabitants are datatypes (which are in turn inhabited by computations and the results of computations). Set
does not belong to itself, so it too exhibits stratification:
Check Set.
Set
: Type
Unlike the previous example, things which quantify over all of Set
do not belong to Set
. This means that Set
exhibits ramified stratification.
Check (forall a:Set, a).
forall a : Set, a
: Type
So, in short, "ramification" in Russell's type hierarchy is embodied today by what Coq calls "predicative" types -- that is, all types except Prop
. If you quantify over a type, the resulting term no longer inhabits that type unless the type was impreciative (and Prop is the only impredicative type).
The higher levels of the Coq universe (Type
) are also ramified, but Coq hides the ramification indices from you unless you ask to see them:
Set Printing Universes.
Check (forall a:Type, Type).
Type (* Top.15 *) -> Type (* Top.16 *)
: Type (* max((Top.15)+1, (Top.16)+1) *)
Think of Top.15
as a variable, like $\alpha_{15}$. Here, Coq is telling you that if you quantify over the $\alpha_{15}^{th}$ universe to produce a result in the $\alpha_{16}^{th}$ universe, the resulting term will fall in the $max(\alpha_{15}+1, \alpha_{16}+1)^{th}$ universe -- which is at least "one level up" from what you're quantifying over.
Just as it was later discovered that Russell's ramification was unnecessary (for logic), it turns out that predicativity is unnecessary for the purely logical portion of CiC (that is, Prop
).