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:
- axiom of extensionality: $x_0=x_1\leftrightarrow\forall x_2\left(x_2\in x_0\leftrightarrow x_2\in x_1\right)$
- reflexivity of equality: $x=x$
- symmetry of equality: $x_0=x_1\rightarrow x_1=x_0$
- transitivity of equality: $x_0=x_1\land x_1=x_2\rightarrow x_0=x_2$
- class comprehension: $x\in\left\{y:\psi\left(y\right)\right\}\leftrightarrow\text{set}\left(x\right)\land\psi\left(x\right)$
- definition of singleton: $\left\{x\right\}=\left\{y:y=x\right\}$
- 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.
Best Answer
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, $\{x_0\}=\{x_1\}$.