Language: Multi-sorted first order logic with equality and membership, where for each natural $n$ we have variables $x\_i^n$ of sort $n$, and for each decidable monotonic strictly increasing sequence of naturals $s$ [with 0 in its domain] ,we have binary relation symbols $=^s, \in^s$ with the following syntatical restrictions: the symbol $\in^s$ can only occur between variables of sort $s\_n$ on the left to variables of sort $s\_{n+1}$ on the right, generally denoted as $x\_i^{s(n)} \in^s x\_j^{s(n+1)}$ [where $s(n)$ is the $n^{th}$ item in sequence $s$]. On the other hand, the symbol $=^s$ can only occur between variables of the same sort, generally denoted as $x\_i^{s(n)} =^s x\_j^{s(n)}$.


Notation: for simplicity we'll only write the type of a variable at quantification.


**Axioms:** [Multi-sorted ID axioms for each sequence $s$] +


**Extensionality:** $ \forall x^{s\_{n+1}} \, \forall y^{s\_{n+1}}: \forall z^{s\_n} \, ( z \in^s x \iff z \in^s y ) \implies x=^sy$


**Comprehension:** $\exists x^{s\_{n+1}} \forall y^{s\_n} (y \in^s x \iff \phi^s(y))$;


where $\phi^s$ only uses $\in^s,=^s$ as predicates, and the sorts of all variables written as items of $s$.



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> Is this equivalent to Tangled Type Theory "$\sf TTT$" of Holmes [see: [Holmes p:11](https://arxiv.org/pdf/1503.01406.pdf), [Holmes p:4-5](https://randall-holmes.github.io/Papers/tangled.pdf)]?
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In the presentation by Holmes there is seeminly one membership and equality relation, unlike here where there is one per type sequence. I was personally thinking of a proof by compactness since every finite fragment of $\sf TTT$ per sequence $s$ is interpretable here, but I'm not that sure?!



Is this theory equivalent to Tangled Type Theory?
It is not. It is very important, very awful, and quite intended that the membership relation between the same two types is the same in different sequences of types.


