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Note 1. Early I posted a related question Set-theoretic tautologies. But the answer did not contain any concrete references to the literature. So I posted this, more precisely formulated question, hoping to receive a concrete reference to the literature. In any case, I am thankful to François G. Dorais and Emil Jeřábek for their answers.

Note 2. Because the 2-element Boolean algebra $\mathbf2$ is isomorphic to the powerset algebra on a singleton $V$, we can use the set-theoretic notations $\emptyset, V, \cup, \cap, C, \subseteq$ instead of normally used in $\mathbf2$ the notations $0,1,+,.,',\leq$.

Let $B$ be a power set Boolean algebra.

Let $P$ be an unquantified formula constructed using variables and the symbols $\emptyset, V$ (the universe), $\cup, \cap, C, \subseteq, =$ (union, intersection, complement, inclusion, equality, respectively) such as, for example, $A \subseteq A \cup X$.

It is well known, that such a formula is valid in $B$ iff it is valid in the 2-element Boolean algebra $\mathbf2$.

So instead of proving $P$ in $B$, we can check that $P$ is true in $\mathbf2$ - by checking $2^n$ possible cases ($n$ is the number of variables in $P$) substituting instead of each variable $\emptyset$ or $V$ and calculating the result, using the obvious rules: $\emptyset \cup \emptyset = \emptyset$, $\emptyset \cup V = V$, $V \cup \emptyset = V$, $V \cup V = V$ and so on for $\cap, C, \subseteq$.

If the result is $\top$ in all cases then $P$ is true in $\mathbf2$ and, therefore, in $B$.

Note that this method also works for some unquantified formulas containing also some propositional connectives - in addition, after calculating all set terms in $P$ we should check that the resulting formula(not containing variables) is true. For example, $A \subseteq B \rightarrow A \cup X \subseteq B \cup X$ is true in $\mathbf2$ and $B$.

But there are some formulas(for example, $X \subseteq Y \lor Y \subseteq X$) that are valid in $\mathbf2$ but not in $B$.

So my question is:

What is the widest class of unquantified formulas in $B$ containing variables, the above set-theoretic symbols and propositional connectives, which are valid simultaneously in $\mathbf2$ and in $B$?

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    $\begingroup$ This sounds like positive logic to me... E.g. I think that if your class of formulae can define negation then it will be too broad for your purposes. $\endgroup$ Commented Jun 18, 2023 at 8:15
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    $\begingroup$ The question is a bit confusing because (IIUC) you actually allow two sets of logical operators: $∩,∪,∁,ø,V$ at the level of sets on the one hand, and on the other hand $∧,∨,¬,⊥,⊤$ at the level of propositions, plus $⊆,=$ which make two sets into a proposition. At least that's what I infer from your examples, because that's not how I had initially understood the sentence. So maybe you should clarify exactly what you allow (by writing the complete set of inductive rules or, at least, by mentioning $∧,∨,¬,⊥,⊤$ from the start). $\endgroup$ Commented Jun 18, 2023 at 8:40
  • $\begingroup$ Alternatively, rewrite sets as predicates in one variable (always the same, $x$), in which case the only connectors allowed are $∧,∨,¬,⊥,⊤$, but quantification in $x$ is allowed in a very specific way, in the form $∀x.(P(x)⇒Q(x))$ (for $P⊆Q$). $\endgroup$ Commented Jun 18, 2023 at 8:42
  • $\begingroup$ @Gro Tsen. I have changed the text, maybe it is more clear now $\endgroup$ Commented Jun 18, 2023 at 16:04

1 Answer 1

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The class of formulas you're looking for contains all quasi-identities: $$a_1 = b_1 \land \cdots \land a_k = b_k \to a = b$$ where $k \geq 0$ and $a,a_1,\ldots,a_k,b,b_1,\ldots,b_k$ are terms formed using $\varnothing,V,{\complement},{\cap},{\cup}$. (We can omit ${\subseteq}$ because $a \subseteq b \iff a \cap b = a$, for example.) I believe these may be exactly the class you're looking for, up to logical equivalence modulo the theory of Boolean algebras.

Some general context: The "well-known fact" from the question is a consequence of Birkhoff's HSP Theorem which says that a nonempty class of algebras is closed under homomorphisms (H), subalgebras (S) and (arbitrary) products (P) if and only if it is axiomatized by identities (i.e., $a = b$ where $a,b$ are terms). There is a generalization of this theorem to quasi-identities which says that a nonempty class of algebras is closed under isomomorphisms (I), subalgebras (S), products (P) and ultraproducts (U) if and only if it is axiomatized by quasi-identities.

A first-order formula $\phi$ (in the language $\varnothing,V,{\complement},{\cap},{\cup},{=}$) with property ISP will have the required property that $\phi$ is valid in $\mathbf{2}$ iff it is valid in every powerset algebra. The forward direction follows from IP since the powerset algebras are isomorphic to powers of $\mathbf{2}$. The reverse direction follows from IS because every nontrivial Boolean algebra contains an isomorphic copy of $\mathbf{2}$.

Note that every first-order formula has property IU by Łoś's Theorem, and every quantifier-free formula has property S. So perhaps the quasi-identities are what you're looking for?


Addendum: What if the powerset algebras are not meant to include the trivial Boolean algebra? Then the above can be modified to include Horn formulas of the form $$a_1 = b_1 \land \cdots \land a_k = b_k \to \bot$$ as well, provided that property P is restricted to nonempty products everywhere.

However, this is not a very substantial extension since $$a_1 = b_1 \land \cdots \land a_k = b_k \to V = \varnothing$$ is a quasi-identity which is equivalent to the above for all nontrivial Boolean algebras.

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  • $\begingroup$ Thank you very much for your answer. I will use it in my [proof-checking system Veda][1]. [1]: github.com/VictorMakarov1/Veda $\endgroup$ Commented Jun 19, 2023 at 4:50
  • $\begingroup$ Awesome! Don't forget to cite this answer if and where it is relevant! $\endgroup$ Commented Jun 19, 2023 at 6:22

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