Are propositional atoms recoverable from this Boolean algebra structure? Consider a set $P$ of propositional atoms, indexed by the positive integers: $p_1, p_2, p_3, ...$. We also have the connectives $\land$, $\vee$, and $\neg$, as well as the parenthesis $($ and $)$. The set of well-formed formulas is defined in the standard way. We define the equivalence relation $E$ on the set of well-formed formulas by identifying logically equivalent well-formed formulas. Let $B$ be the set of equivalence classes under this equivalence relation. We can form the structure $(B;0,1,\land,\vee, \neg)$, where $0$ is the set of all contradictions, $1$ is the set of all tautologies, and $\land, \vee, \neg$ are defined by passing to representatives. My question is, is the set of equivalence classes of all propositional atoms definable in that structure? I conjecture that it is not. In fact, I make the stronger conjecture that no non-empty subset of propositional atoms is definable. By definable, I mean definable without parameters. Are either or both of these conjectures true?
 A: No, it is not.
The structure you have defined is the free Boolean algebra on $\mathbb{N}$. This means there is a function $n \mapsto p_n : \mathbb{N} \to B$ such that for all Boolean algebras $B’$ and all functions $f : \mathbb{Ν} \to B’$, there exists a unique Boolean algebra homomorphism $g : B \to B’$ such that $\forall n (g(p_n) = f(n))$.
In particular, let $g : B \to B$ be the unique homomorphism sending each $p_i$ to $\neg p_i$. Note that $g(g(p_i)) = g(\neg p_i) = \neg g(p_i) = \neg \neg p_i = p_i$. That is, $g \circ g$ is the unique homomorphism sending $p_i$ to $p_i$. The identity map is also a homomorphism sending each $p_i$ to $p_i$, so $g \circ g$ is the identity.
In particular, then, $g$ is a Boolean algebra isomorphism (with itself as its inverse). Therefore, anything which could be used to define $\{p_i \mid i \in \mathbb{N}\}$ purely in terms of the Boolean algebra structure of $B$ would also define $\{\neg p_i \mid i \in \mathbb{N}\}$, which is a different set.
The situation is analogous to the standard basis unit vectors in the vector space $\mathbb{R}^n$. There is no way to define what it means to be a standard basis unit vector purely in terms of vector space structure because these vectors are not necessarily preserved by isomorphism - in particular, the isomorphism $x \mapsto -x$.
A: The Boolean algebra you describe is the free Boolean algebra on countably many generators (the equivalence classes of the propositional atoms).
The easiest way to prove your two conjectures is to note that the homomorphism $B\to B$ induced by $p_i\to \lnot p_i$ for all $i$ is an automorphism (it is its own inverse). Since it maps the set of propositional atoms to a disjoint set (their negations), no non-empty subset of the set of propositional atoms is definable.
$B$ can also be characterized as the unique (up to isomorphism) countable atomless Boolean algebra. It is the Fraïssé limit of the class of finite Boolean algebras. Therefore it is homogeneous and admits quantifier elimination. More concretely, any isomorphism between finite subalgebras of $B$ extends to an automorphism of $B$.
Since every singleton other than $0$ and $1$ generates the unique Boolean algebra with $4$ elements, any two such elements are conjugate by an automorphism. Hence the automorphism orbits of $B$ are $\{0\}$, $\{1\}$, and $B\setminus\{0,1\}$. Thus there are only $8$ definable subsets of $B$: the unions of these three orbits.
Of course there are more complicated definable subsets of $B^n$ for $n>1$, or if we allow parameters in our definitions.
