Consider a Boolean set algebra on a set $\Omega$. Let $\sigma$ be a set function on $\Omega$ such that for all $m\in \Omega$ $\sigma(m)\subset \Omega$. The operator $\square_\sigma$ is defined by $\square_\sigma P=\{m:\sigma(m)\subset P\}$ for all $P\subset \Omega$. It has the following properties:

1. $\square_\sigma \Omega= \Omega$
2. $\square_\sigma(P\cap Q)=\square_\sigma P \cap \square_\sigma Q$
3. $\square_\sigma(P\to Q)\to (\square_\sigma P\to \square_\sigma Q)=\Omega$

If for all $m\in \Omega$ $m\in \sigma(m)$, then

4. $\square_\sigma \emptyset = \emptyset$
5. $\square_\sigma P \to P= \Omega$ $\quad (\square_\sigma P\subset P)$
6. if $\square_\sigma P=\Omega$, then $P=\Omega$

and if for all $m\in \Omega$ and for all $k \in \sigma(m)$ $\sigma(k)\subset \sigma(m)$, then

7. $\square_\sigma \square_\sigma P= \square_\sigma P$.

Is this definition of the modal operator $\square_\sigma$ general?