# Question about infinitary logic

Let $\mathcal{L}$ be the language based on countably many sentence letters $p_i$, $\neg$, and infinite conjunction $\bigwedge$. Say that $p_i$ is an essential constituent of an $\mathcal{L}$-sentence $\varphi$ iff $p_i$ occurs in every $\mathcal{L}$-sentence $\psi$ s.t. $\vDash \varphi \leftrightarrow \psi$. Say that $\varphi$ is qualitative iff it has no essential constituents.

Question 1: Must any $\mathcal{L}$-sentence built from qualitative $\mathcal{L}$-sentences using $\neg$ and $\bigwedge$ be qualitative? (Note: We may form uncountable conjunctions.)

Question 2: If "yes", what about in languages with uncountably many sentence letters, or in infinitary first-order languages (now requiring essential constituents to be individual constants)?

Question 3: Is there a literature on this sort of thing? (Maybe in the literature on interpolation in infinitary languages?)

(Observation: The answer to Q1 is obviously "yes" in the finitary fragment of $\mathcal{L}$, since in that fragment only validities and their negations are qualitative. But that's not true in $\mathcal{L}$. For example, neither $\bigvee_{0\leq i}\bigwedge_{i\leq j}p_j$ nor its negation is a validity, but it is qualitative, since for all $n$, $p_n$ does not occur in $\bigvee_{n+1\leq i}\bigwedge_{i\leq j}p_j$ yet $\vDash \bigvee_{0\leq i}\bigwedge_{i\leq j}p_j \leftrightarrow \bigvee_{n+1\leq i}\bigwedge_{i\leq j}p_j$.)

Answer to Q1: Yes. For every $p_i$ we can show that it is not an essential constituent in the large sentence, simply by simultaneously replacing each of the qualitative subsentences by an equivalent that doesn't mention $p_i$. The truth-value evaluation for the large sentence cannot be affected by that.