Are propositional atoms recoverable from this Boolean algebra structure

Question

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?

Answer 1

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 $x \mapsto -x$.