In how many ways can the power set of an $n$-element set be divided into two disjoint subsets: the sets $I$ and $H$, such that the following conditions are simultaneously satisfied?

a) for any $a, b \in I$, $a \cup b \in I$ and $a \cap b \in I$

b) for any $a, b \in H$, $a \cup b \in H$ and $a \cap b \in H$;

c) for any $a \in I, b \in H$, $a \cup b \in I$ and $a \cap b \in H$.

(The power set of a set is the set of all its subsets.)
Let the set with $n$ elements be denoted by $S$, its elements being $x_{1}, \ldots, x_{n}$.

It is easily verifiable that the two decompositions where all subsets are placed into $H$ or $I$, meaning that one of $H$ or $I$ is empty, are valid. We will call these trivial decompositions. In the following, we will precisely describe the non-trivial decompositions, where neither $H$ nor $I$ is empty.

First, we will show that in a non-trivial decomposition, $\emptyset \in H$ and $S \in I$. Let $h \in H$ and $i \in I$ be any two sets; then $\emptyset=\emptyset \cap h \in H$ and $S=S \cup i \in I$ due to properties a), b), and c).

Suppose $n \geq 1$, and consider now the single-element subsets $\left\{x_{1}\right\}, \ldots,\left\{x_{n}\right\}$. We claim that exactly one of these belongs to $I$. If all single-element sets belonged to $H$, then by property c), the set $\left(\left\{x_{1}\right\} \cup\left\{x_{2}\right\}\right) \cup \cdots \cup\left\{x_{n}\right\}=S$ would also be in $H$, which is a contradiction. Therefore, there is at least one single-element set that belongs to $I$. If at least two single-element sets belonged to $I$, then by property a), their intersection, $\emptyset$, would also be in $I$, which is also a contradiction.

Let $\left\{x_{j}\right\}$ be the single-element set that belongs to $I$. Based on properties a) and b), we can conclude that every subset $r$ that contains $x_{j}$ belongs to $I$; indeed, $r=r \cup\left\{x_{j}\right\}$.

Conversely, if a subset $r$ does not contain $x_{j}$, then $r \in I$ is not possible; if $r \in I$, then by property a), $r \cap\left\{x_{j}\right\}=\emptyset$ would also be in $I$. Therefore, we have that a subset $r$ belongs to $I$ if and only if $x_{j} \in r$. The number of non-trivial decompositions is thus the number of possible $x_{j}$, which is $n$.

In the case $n=0$, i.e., $S=\emptyset$, the two trivial decompositions are the same: $I=H=\emptyset$ and there are no other decompositions; the number of all decompositions is therefore 1. If $n=1$, then the two trivial decompositions are different, and there are no other decompositions, so the number of decompositions is 2. If $n>1$, then the two trivial decompositions are different, and the number of all decompositions is $n+2$.