Q: Is every formula of $L_\Omega$ equivalent to a formula of $L_1$ modulo $T_1$?
The notation comes from the following question: Is the following theory countably axiomatizable?
Edit: I mean $T_\Omega$.
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Sign up to join this communityQ: Is every formula of $L_\Omega$ equivalent to a formula of $L_1$ modulo $T_1$?
The notation comes from the following question: Is the following theory countably axiomatizable?
Edit: I mean $T_\Omega$.
Noah is correct that over the theory $T_1$, we don't know about the new constants. But meanwhile, over the final theory $T_\Omega$, or what I would call $T_{\omega_1}$, every formula in the extended language is equivalent to a formula in the language of (class) set theory itself. You have added constants $c_\phi$ and axioms $\forall x\ (x\in c_\phi\iff \phi(x))$, where $\phi$ uses only earlier-mentioned constants. So by induction, any instance of $x\in c_\phi$ that appears in a formula can be systematically eliminated by this means, and so in fact we don't need any of these new constants.
So over your final theory, every formula is equivalant to an assertion in the second-order language of set theory.
No, they are not. There are formulas in $L_2$, for example, involving constant symbols in $L_2$ but not in $L_1$. $T_1$ has nothing to say about those constant symbols, so no formula involving such a constant symbol (in a nontrivial manner) can be $T_1$-equivalent to an $L_1$ formula.
The sense in which $T_2$ is "no more than" $T_1$ is interpretability (and conservativity) - any model of $T_1$ interprets an expansion of itself which is a model of $T_2$.