= Solution
Take $S_1$ to consist of the three usual <first-order sentences>
$$
\forall x\ E(x,x),\qquad
\forall x\forall y\,(E(x,y)\to E(y,x)),\qquad
\forall x\forall y\forall z\,((E(x,y)\wedge E(y,z))\to E(x,z)).
$$
These assert <reflexivity>, <symmetry> and <transitivity> and therefore define exactly the <equivalence relations>.
For $S_2$, adjoin to $S_1$ the single <first-order sentence>
$$
\forall x\exists y\,[y\ne x\wedge E(x,y)\wedge
\forall z\,(E(x,z)\to(z=x\vee z=y))].
$$
Its witness $y$ gives a second member of each class, while its final universal clause excludes all others. Hence every class has exactly two members, and conversely such a relation satisfies it. This remains a <finite set> of <first-order sentences>.
For each positive <integer> $n$, write
$$
\theta_n(x)=\exists y_1\cdots\exists y_n
\left(\bigwedge_{i=1}^n E(x,y_i)\wedge
\bigwedge_{1\le i<j\le n}y_i\ne y_j\right).
$$
Then
$$
\boxed{S_3=S_1\cup\{\forall x\,\theta_n(x):n\ge1\}.}
$$
Every class in a <first-order model> has at least $n$ elements for every $n$, so cannot be finite. Conversely every infinite <equivalence class> supplies each finite collection of distinct witnesses. The displayed infinite scheme therefore has exactly the requested <first-order models>.
Back to article page
