Hey friends of maths,
I am trying to find an example that fulfills the requirements mentioned above, more precisely:
I'd appreciate your help with finding two (or better more than 200) non-isomorphic finite structures where the axiom $\forall x \exists P \forall z (z \in P \leftrightarrow z = x)$ does hold, and also, if possible, where the axioms $\forall A \forall B ([\forall x: x \in A \Leftrightarrow x \in B] \Rightarrow A = B)$ and $\exists N:(\forall x: x \in N \Leftrightarrow x \neq x)$ - do hold.
This will help: it is impossible to satisfy both the axiom of the empty set and the singleton axiom using a finite sized structure:
First, by axiom of the empty set, you need an object, call it $d_0$, and for whatever other objects you have in the domain, those objects cannot stand in the relation as denoted by $\in$ (let's call this relarion $R$) with it.
Second, by the axiom of the singleton, there needs to be an object such that $d_0$ is the one and only object standing in the relation $R$ with it. For the reason just stated earlier, this cannot be $d_0$ itself, and so we need a new object $d_1$. So, we have $(d_0,d_1)\in R$ ... and there cannot be any other objects standing in relation $R$ to $d_1$.
OK, but then by the axiom of the singleton again, there needs to be some objects such that $d_1$ is the one and only object standing in relation $R$ with it ... and this cannot be $d_0$ (since nothing stands in relarion $R$ to $d_0$), not can it be $d_1$, since there cannot be any object other than $d_0$ standing in relation $R$ to $d_1$. So, we need to have anew object, $d_2$, and we need that $(d_1,d_2)\in R$, and we need to make sure that nothing but $d_1$ stands in the relation $R$ to $d_2$.
... ok, I think you see where this is going ... you indefinitely need to keep adding one more object $d_{i+1}$ to the already existing objects $d_0$ through $d_i$, because in order to satisfy the singleton axiom for $d_i$ we need some object that $d_i$ stands in the relation $R$ to, and by the axiom of the empty set that object cannot be $d_0$ and since we have $(d_j,d_{j+1})\in R$ for all $0\le j \le i-1$, all objects $d_1$ through $d_i$ already have some object other than $d_i$ standing in the relation $R$ to them and are not allowed to have any other object stand in relation $R$ to them.