A question was just asked here about proving
$$A⊆(B∪C)⟺A\setminus C⊆B.$$
We can prove this statement directly, using the concepts of first-order logic.
"Suppose $x \in A \setminus C$ and that $A⊆(B∪C).$ We will show $x \in B$..."
etc.
Instead of doing that, is there a way to leverage the connection between Boolean algebras and Propositional Logic to transform the statement of interest into a statement in the language of Propositional Logic?
If so, we can probably prove the new statement using more elementary means.
Yes we can. In fact, it is rather nice to see how these things play out.
We replace the sets $A,B,C$ with the predicates $\sf A,B,C$, where $\sf A$ is to be thought of as "$\in A$". Accordingly, we need to adjust the statement to prove to:
$$(\sf A \implies (B \lor C))\iff ((A \land \neg C) \implies B)$$
which is essentially not but applying the correspondence between PropLog and Boolean algebras, after restricting to the Boolean subalgebra of $\mathcal P(U)$ (where $U$ is the universe of discourse for the original statement) determined by $A,B,C$.
We can now solve this using truth tables, more properly known as truth valuations. It is here that the connection with the first-order case becomes apparent.
For suppose we had a valuation $x$, with according truth values $x({\sf A}),x({\sf B})$ and $x({\sf C})$. Now simply think of $x({\sf A})$ as $x \in A$.
So we see that the first-order reasoning in this case can be argued to actually be zeroth-order reasoning in disguise: we just prefer to call our truth valuations (arbitrary) elements.