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Comments on On Tarski-style universes in type theory

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On Tarski-style universes in type theory

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I'm looking at this note about universes in type theory.

For Tarski-style universes (page 2), as far as I understand, the "$a$" in $a:U_i$ are formal names of types, whereas the "$T_i(a)$" in $T_i(a) \ type$ are actual types. What I don't understand is the purpose of "$u_i:U_{i+1}$" and "$T_{i+1}(u_i)=U_i$". Is there any intuitive explanation on why we have these two "axioms" (or whatever)?

Further, it is said in the remark on page 3 that Tarski-style universes can be formulated in LF:

$U_i: \textbf{Type}; \ T_i: (U_i)\textbf{Type};\ u_i:U_{i+1};\ T_{i+1}(u_i)=U_i : \textbf{Type}$

But how is this related to $El(A)$ from the LF mentioned in my previous question? Is there any relationship between $El(-)$ and $T_j(-)$?

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I'm a bit confused about what you're confused about for the first part. In a Russell-style universe system, we'd have a hierarchy of universes with $\mathcal U_i : \mathcal U_{i+1}$, i.e. the $i$th universe is a type in the $(i+1)$th universe. As you state, for a Tarski-style universe system, $\mathcal U_i$ is a type of "codes" for types in the $i$th universe and $T_i(a)$ for $a : \mathcal U_i$ is the "actual" type, i.e. a thing that can be to the right of a "$:$". Thus, there should be a code $u_i$ in $\mathcal U_{i+1}$ that represents the fact that $\mathcal U_i$ is a type in the $(i+1)$th universe. That is, $u_i : \mathcal U_{i+1}$ and $T_{i+1}(u_i) = \mathcal U_i$. This is just directly how you say the Russell-style $\mathcal U_i : \mathcal U_{i+1}$ in a Tarski-style universe.

As for how $El$ relates to $T_i$, $El$ and $T_0$ are the same thing where LF has only one universe, namely $\mathsf{Type}$. So $\mathcal U_0$ correspond to $\mathsf{Type}$, $T_0$ corresponds to $El$, and $type$ corresponds to $kind$. To be clear, this is treating LF and this logic with Tarski-style universes on the same footing. If we wanted to formalize this logic in LF, then the $El$ of LF as the meta-language and the $T_0$ of the object language would be different kinds of things.

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I think it would be easier for me to understand the second part of your answer if we consider a small... (4 comments)
I think it would be easier for me to understand the second part of your answer if we consider a small...
user205‭ wrote over 1 year ago

I think it would be easier for me to understand the second part of your answer if we consider a small concrete example, so I asked a separate question about the two universes $\ast$ and $\square$ from CoC.

One question that I didn't include to my new post: if we use Tarski-style universes and if we want to get a consistent logical theory, can the length of the chain $U_i: U_{i+1} : \dots$ be finite? I'm thinking about CoC, which is consistent, and where we have two universes $\ast$ and $\square$ such that $\ast:\square$. As far as I understand, for Russel-style universes, an infinite chain is needed to avoid things like $U:U$.

Derek Elkins‭ wrote over 1 year ago

Sure it can be finite. It is for LF. The top-level universe is simply not a term of any other universe. The universe being Tarki-style or Russell-style doesn't change that. (This is why LF doesn't have a code for $\mathsf{Type}$ and a corresponding $El(\mathsf{type})=\mathsf{Type}$ equation. If we did add those, we'd almost certainly run into inconsistency, because those would effectively be saying $\mathsf{Type}:\mathsf{Type}$ from a Russell-style perspective. So we just don't add those.)

The benefit of an infinite hierarchy of universes (besides power) is that we can uniformly say that every term is classified by some type. This avoids the need for a parallel set of rules for terms that are not classified by some type. You can see this happening in LF where we need a separate $\Gamma\vdash K\ kind$ judgement distinct from $\Gamma\vdash k:K$ judgement with a lot of parallel rules.

user205‭ wrote over 1 year ago

Let's consider Calculus of Constructions again. In it, there isn't an infinite hierarchy of universes, there are only two universes if my understanding is correct. So I was wondering if there is any example of a term in Calculus of Constructions that wouldn't be "classified by some type"? And what is the "parallel set of rules" in the case of CoC? I only know of these rules, is there anything else?

Derek Elkins‭ wrote over 1 year ago · edited over 1 year ago

If you treat $\square$ as a term (which is probably not the case), then it is a term that is not classified by a type. If you don't treat it as a term (and in the original description there was no $\square$), then $*$ is a term that isn't classified by a type, and $(-)\vdash(-):\square$ should be treated as a binary relation and not a special case of the trinary $(-)\vdash(-):(-)$. The parallel rules in your description are encoded by the $s$ (meta-)variables which can be either $*$ or $\square$. In the original description, these parallel rules are more explicit, e.g. there's $$\dfrac{\Gamma[x:M]\vdash\Delta}{\Gamma\vdash[x:M]\Delta}\quad\text{and}\quad\dfrac{\Gamma[x:M_1]\vdash M_2:*}{\Gamma\vdash[x:M_1]M_2:*}$$ The rules you list leverage presentational innovations from Pure Type Systems which came after and were, in part, inspired by the original description of the Calculus of Constructions.