I first want to proof that the intersection of two inductively defined sets is again inductive. We have $A$ which is a set of elementary elements and a set of functions $F$ where each function has arity $k \geq 1$. The set $S$ is the smallest set inductively defined through $A$ and $F$ for which the follwing is true:
(1) $A \subseteq S$
(2) $f \in F$ and $f$ has arity $k$ and $s_1,...,s_k$ are elements of $S$. Then $f(s_1,...,s_k)$ is also in $S$.
Assume that $S_1$ and $S_2$ are both sets for which (1) and (2) are true. I want to show that the rules (1) and (2) are also true for the intersection $S_1 \cap S_2 = S$.
I am not even sure where to start. What does it even mean that $S_1$ and $S_2$ are sets for which both (1) and (2) are true? Do I know that $A_1 \subseteq S_1$ and $A_2 \subseteq S_2$ or do I know that $A \subseteq S_1$ and $A \subseteq S_2$ or is it irrelevant in this situation?
Let's assume that $A \subseteq S_1$ and $A \subseteq S_2$. Therefore we also know that $A \subseteq S$ by definition of the intersection. So (1) is true for $S$.
Is this even correct until now? I think if I wanted to show that (2) is true for $S$, I would take an arbitrary function $f$ with arity $k \geq 1$ and $s_1,...,s_k$ elements and show that $f(s_1,...,s_k) \in S$.