Boolean algebra

Main example: the subsets of a set form a Boolean algebra. To see why, let's define a set $S$ and its power set $\mathcal{P}(S)$, which is the set of all subsets of $S$. We can then define two operations on $\mathcal{P}(S)$, union ($\cup$) and intersection ($\cap$), as well as a complement operation (${}^{\prime }$) that takes the complement of a set with respect to $S$.
Formally, we can define the Boolean algebra $\mathcal{B}(S)$ as the tuple $(\mathcal{P}(S),\cup ,\cap {,}^{\prime },\mathrm{\varnothing },S)$. We can then verify that these operations satisfy the axioms of a Boolean algebra:

1. Commutativity of $\cup$ and $\cap$:

2. Associativity of $\cup$ and $\cap$:

3. Distributivity of $\cup$ over $\cap$ and $\cap$ over $\cup$:

4. Identity elements:

6. Complement:

7. Double complement:

This Boolean algebra formalize the usual Logic, with intersection and union in the place of "and" and "or", and the subsets being "propositions".
In the context of sets, the "implies" operator corresponds to set inclusion. That is, if $A$ and $B$ are sets, then $A\to B$ means that if an element is in $A$, then it must also be in $B$. Symbolically, $A\to B$ is equivalent to $x\in A\to x\in B$. Alternatively, we can also define $A\to B$ as the set of all elements that are either not in $A$ or that are in both $A$ and $B$. Symbolically, $A\to B=(A^c\cup (A\cap B))=A^c\cup B$, where $A^c$ is the complement of $A$ (that is, the set of all elements not in $A$).

But even more, what in Logic is considered a "verifiable proposition" can be understood as open sets, endowing in this way to the set $S$ with a topology. See this answer or this other and this blog entry. Maybe ideas in the blog post are related to new logic for Quantum Mechanics.