Is it possible to build set theory on first-order logic *without equality*?


For example, how could one show that if $x\_0=x\_1$ then $\left\{x\_0\}=\{x\_1\right\}$, where $x\_0$ and $x\_1$ are two sets? And the other way around? In my opinion, it is impossible using only the following facts:


1. axiom of extensionality: $x\_0=x\_1\leftrightarrow\forall x\_2\left(x\_2\in x\_0\leftrightarrow x\_2\in x\_1\right)$
2. reflexivity of equality: $x=x$
3. symmetry of equality: $x\_0=x\_1\rightarrow x\_1=x\_0$
4. transitivity of equality: $x\_0=x\_1\land x\_1=x\_2\rightarrow x\_0=x\_2$
5. class comprehension: $x\in\left\{y:\psi\left(y\right)\right\}\leftrightarrow\text{set}\left(x\right)\land\psi\left(x\right)$
6. definition of singleton: $\left\{x\right\}=\left\{y:y=x\right\}$
7. definition of sethood: $\text{set}\left(x\_0\right)\leftrightarrow\exists x\_1\left(x\_0\in x\_1\right)$


I hope I didn't forget any fact. Also, it would be better if I point out I'm using Morse-Kelley set theory.


If I use first-order logic *with equality*, I could prove the first part of the theorem because for any term $t\_0$ and $t\_1$, if $t\_0=t\_1$ then $f\left(t\_0\right)=f\left(t\_1\right)$ ($f$ is a unary logical function here). But how can I prove the other way around?


Thanks.



First-order logic without equality and set theory
The converse direction fails because, in Morse-Kelley set theory plus your facts, $\{x\}$ is empty whenever $x$ is a proper class. So $\{x\_0\}=\{x\_1\}$ for any two proper classes.


The forward direction, on the other hand, seems to follow immediately from the facts you listed. If $x\_0=x\_1$ then, since facts 2, 3, and 4 make equality an equivalence relation, any $y$ will be equal to $x\_0$ iff it is equal to $x\_1$. So by facts 5 and 6, $\{x\_0\}$ and $\{x\_1\}$ have the same members. Then by fact 1, $\{x\_0\}=\{x\_1\}$. 


