What is the set theory synonymous with this order-set theory?

Let $$T$$ be a theory written in Mono-sorted first order logic with equality, with extralogical primitives: $$<, \in$$.

Define: $$x \leq y \iff x < y \lor x=y$$

Axioms:

• $$\textbf{Well ordering: }\\\textit{Transitive:} \ x < y \land y < z \to x < z \\ \textit{Connective:} \ x \neq y \leftrightarrow (x < y \lor y < x) \\ \textit{Well founded:} \ \exists n \in x \to \exists n \in x \forall m \in x (n \leq m)$$

• $$\textbf{Finiteness: }\exists n \in x \to \exists n \in x \forall m \in x (m \leq n)$$

• $$\textbf{Sets: } \forall n \exists! x \forall m (m \in x \leftrightarrow m \leq n \land \phi)$$, if $$x$$ is not free in formula $$\phi$$.

• $$\textbf{Closure: } x \in y \to x < y$$

• $$\textbf{Emergence: }\max ( x \setminus y) < \max (y \setminus x) \to x < y$$

Where, $$\max x$$ is the maximal element of $$x$$ with respect to relation $$<$$; and $$x \setminus y = \{m \in x \mid m \not \in y\}$$.

Theory $$T$$ is an example of an interwoven theme of order and set theory. It is synonymous with $$\sf PA$$. [see here]

If we remove Finiteness, and add the existence of a limit number, that is:

• $$\textbf{Infinity: } \exists l \neq \emptyset: \forall x < l \exists y: x < y < l$$

, call the resultant theory $$T - {\sf Fin + I}"$$.

What would be the set theory that is synonymous to $$T - {\sf Fin + I}$$?

My guess is that it would be $$\sf Z + Ranks$$ plus adding a bijection $$j$$ to the language of $$\sf Z$$ from sets to von Neumann ordinals in a manner such that sets are assigned higher ordinals than their elements. The Ranks axiom is $$(\forall \alpha \exists V_\alpha)$$ where $$V_\alpha = \{x \in j^{-1}(\beta) \mid \beta < \alpha \}$$. So, the theory would prove Global Choice, Transitive closures, Ranks, etc...

[EDIT:] The above theory had been proved inconsistent as shown in the accepted answer. A possible salvage along the original intentions is to restrict axiom of Emergence to apply to natural numbers (i.e. objects prior to the first infinite number). We call $$T$$ with this restriction as $$T^r$$. We add the following axiom:

• $$\textbf {Respective: } \forall n \in x (m > n) \land m \in y \to x < y$$.

So, the question is re-stated in connection with theory $$T^r - {\sf Fin + I + Respective}$$.

• How is $\max x$ defined in the axiom of Emergence in the resulting theory? Without Finiteness it is entirely possible that a set exists with no maximum element, and I think Infinity implies one definitely exists. Jan 10 at 3:41
• @pastebee, "$\max x < \max y$" is defined as: $\exists a \exists b \, (a=\max x \land b= \max y \land a < b)$ Jan 10 at 5:47

We can define $$0$$ and $$S(n)$$ as in my previous answer - $$0$$ is the smallest number, $$S(n)$$ is the smallest number greater than $$n$$. My proof that these exist did not use Finiteness.
Define $$n$$ to be a natural number if, for any set $$x$$, if $$0 \in x$$, and for all $$k \in x$$, $$S(k) \in x$$, then $$n \in x$$. For a limit number $$l$$, all natural numbers are less than $$l$$: $$0$$ is clearly less than $$l$$, and for any $$n < l$$, there exists $$y$$ such that $$n < y < l$$, and therefore $$S(n) \leq y < l$$. So the set of natural numbers exists, and I will write it as $$\mathbb{N}$$.
If $$n, m$$ are natural numbers and $$n < m$$, then $$\mathbb{N} \setminus \{m\} < \mathbb{N} \setminus \{n\}$$, by Emergence: $$(\mathbb{N}\setminus\{m\})\setminus(\mathbb{N}\setminus\{n\}) = \{n\}$$, and similarly $$(\mathbb{N}\setminus\{n\})\setminus(\mathbb{N}\setminus\{m\}) = \{m\}$$, and $$\max\{n\} < \max\{m\}$$.
For all $$n \in \mathbb{N}$$, $$\mathbb{N} \setminus \{n\} \leq \mathbb{N} \setminus \{0\}$$, so the set $$\{\mathbb{N} \setminus \{n\} : n \in \mathbb{N}\}$$ exists. For any element $$\mathbb{N} \setminus \{n\}$$ of this set, $$\mathbb{N} \setminus \{S(n)\}$$ is a smaller element, so this set has no minimum. This contradicts well-foundedness, so $$T - {\sf Fin + I}$$ is inconsistent.