Let $S$ be the set of all n-ary functions on $\{0,1\}$ for all n, including the 0-ary functions. Let $F$ be a finite subset of $S$. Consider a countably infinite set of propositional constants $PROP$. Now, we define recursively a language on $PROP \cup F$, by stipulating that members of $PROP$ are in the language, and if $f$ is an n-ary function in $F$ and $x_1,...,x_n$ are already in the language, then the concatenation $fx_1,...x_n$ is also in the language.
We can now form a consequence relation on this language, by saying that a member $w$ is implied by a set $W$ iff there is no boolean valuation that simultaneously makes all members of $W$ true but makes $w$ false.
Although the consequence relation has infinitely many members, we can sometimes use a finite generating set. For example, considering the set that only contains conjunction, I believe this is a finite generating set: 1. $\{p_1,p_2\}$ $\vdash$ $p_1 \land p_2$ 2. $\{p_1 \land p_2\}$ $\vdash$ $p_1$ 3.$\{p_1 \land p_2\}$ $\vdash$ $p_2$.
Now that I have set the background theory for my question, my question is, given any finite set of boolean connectives, does it always have a finite generating set? If so, what is the proof.