I Have transferred this posting from Mathoverflow, because I think it is not research level, though it is not too elementary I suppose.
Define $\mathcal P^\equiv (A)$, the stratified power set of set $A$, as the minimal set of subsets of $A$ that has among its elements all subsets of $A$ containing less than three elements, and the unordered intersection relation set over $A$, that is closed under Complements relative to $A$, Set union, and Unordered Composition. Formally that is: $$ \mathcal P^\equiv(A) = \{a \subseteq A| \, \forall X: \mathcal P_{\leq 2}(A) \subset X \land \\ \Pi^*A \in X \land \\ \forall c,d \in X \, (A \backslash c \in X, \bigcup c \in X, c \circ^* d \in X ) \\\to \\ a \in X \}$$
Where: $ \Pi^* A = \{\{x,y\} \in A: x \cap y \neq \emptyset \}$
$c \circ^* d = \{\{x,z\}: \, \exists y \, (\{x,y\} \in c \land \{y,z\} \in d )\}$
My question is that, working in ZF or one of its known extensions, is it possible to have a set $A$ such that: $$|\mathcal P^\equiv(A)| = |A|$$ esepcially when $A$ is transitive and closed under pairing.
My point is that the set $\mathcal P_{\leq 2} (A) \cup \{\Pi^* A\}$ can be of the same cardinality of $A$ for an infinite $A$, so we can begin with it, then add the needed closures successively.
Here, I'll present a plan for that and my question would be specifically on where this fails, should it fail?
first we define $S_0(A) = \mathcal P_{\leq 2} (A) \cup \{\Pi^* A\}$
Now define $S_{i+1}(A) = CL^{[\backslash,\bigcup,\circ^*]}(S_i(A))$
where: $$CL ^{[\backslash,\bigcup,\circ^*]}(S)= \{x \subseteq A: \exists k,l \in S (\\x= (\bigcup S)\backslash k \lor \\ x= \bigcup k \lor \\ x= l \circ^* k) \}$$
Then by replacement take $$\mathcal P^\equiv(A)=\bigcup^{n \in \mathbb N} S_n (A)$$
So clearly $\mathcal P^\equiv(A)$ is the countable union of sets of the same cardinality of $A$, so if $A$ is infinite, and given choice, then that union is always of equal cardinality with $A$.