Solution
= Solution
Suppose the <finitary set of formulae> $S$ is equivalent to the finite set
$$
F=\{f_1,\ldots,f_m\}.
$$
For each $f_j$, the relation $S\vdash f_j$ has a finite proof, so it uses a finite set $R_j\subseteq S$ of assumptions. Put
$$
R=\bigcup_{j=1}^mR_j.
$$
This is a finite subset of $S$, and $R\vdash f_j$ for every $j$, hence $R\vdash F$. Since equivalence also gives $F\vdash S$, transitivity of deduction yields
$$
\boxed{R\vdash S}.
$$