As far as I can see, they only employ different symbols but they operate in the same way. Am I missing something?
I wanted to write "Boolean logic" in the tag box but a message came up saying that if I wanted to write Boolean logic I should better write propositional logic. Doesn't this confirm my suspicions?
you can see :
In this chapter we give a brief introduction to the theory of Boolean algebras. As we shall see, these are algebraic structures which stand in an intimate relationship to [classical] sentential logics [emphasis added]. They will also form a source for some of the applications of logic which we shall give later.
See page 158 :
[regarding] the correspondence between Boolean algebras and sentential logics [...] We shall see that there is a full correspondence between these two kinds of mathematical objects.
Finally, see page 160 :
the following theorem, which is another kind of completeness theorem for Boolean algebras. [...] Hence we may say that the theories of Boolean algebras and of sentential logics are equivalent, in some sense.
You can see also :
a Boolean algebra is a complemented distributive lattice with at least two elements.
Page 40 :
we impose the structure of a boolean algebra on the set $F$ of formulas of [sentential calculus] $SC$ [...] by first defining the relation $\equiv$ on $F$ by :
$$\phi \equiv \psi \ \text {iff} \ \vdash \phi \to \psi \ \text {and} \ \vdash \psi \to \phi.$$
If $\phi$ is a formula in $F$, we let $|\phi|$ the equivalence class under the relation $\equiv$ to which it belongs. Thus
$$|\phi| = \{ \psi \in F : \phi \equiv \psi \}.$$
We let
$$F/\equiv = \{ |\phi| : \phi \in F \}.$$
Now we can define the relation $\le$ on $F/\equiv$ by
$$|\phi| \le |\psi| \ \text {iff} \ \vdash \phi \to \psi.$$
Thus, we have that :
$\mathcal A = \langle F/\equiv \le \rangle$ is a complemented distriutive lattice, that is a Boolean algebra.
