$\equiv$ is an equivalence relation on $P(\mathbb{N})$ ($P(A)$ denotes the power set of the set $A$) defined as $$A\equiv B \iff A,B\subseteq \mathbb{N} \land (\forall X\subseteq\mathbb{N})[A\cup X=\mathbb{N}\iff B\cup X=\mathbb{N}]$$ Is $P(\mathbb{N})/\equiv$ a finite set (the quotient set of $P(\mathbb{N})$ by $\equiv$; so the set of all equivalence classes in $P(\mathbb{N})$ with respect to the relation $\equiv$)?
Note: I have a feeling that the definition of the relation could be simplified but I don't see how. I am not sure what exactly $(\forall X\subseteq\mathbb{N})[A\cup X=\mathbb{N}\iff B\cup X=\mathbb{N}]$ means for $A$ and $B$.
First I am trying to see why the given relation is actually an equivalence relation.
I think it's obvious that $A\equiv A \text{ } \forall A\in P(\mathbb{N})$ as $$A\cup X=\mathbb{N}$$ and $$(A=)B\cup X=\mathbb{N}$$ are the same thing, so of course they are equivalent.
If $A\equiv B,$ then $B\equiv A$ as $A$ and $B$ participate symetrically in the given equivalence. So we would just swap them and get $B\cup X=\mathbb{N} \iff A\cup X=\mathbb{N}$ which gives exactly $B\equiv A$. Note that we started by assuming that $A$ and $B$ are in $\equiv$.
I am not so sure how to check if the relation is transitive though. And I also have no idea how to find the classes of equivalence. Thanks!