MOD(S
\cup \{\neg \hat{a}\})$, and vice versa. Since $MOD(S_1 \cup S_2) = MOD(S_1) \cap MOD(S_2)$, to every model $m \in MOD(S_1 \cup S_2)$ corresponds $(m \cup \{\hat{a}\}) \in MN(S \cup \{\hat{a}\})$ and $(m \cup \{\neg \hat{a}\}) \in MN(S \cup \{\neg \hat{a}\})$, and vice versa. Hence, $N_2 = |MN(S \cup \{\hat{a}\})| = |MN(S \cup \{\neg\hat{a}\})|$. So line 5 of the algorithm verifies whether $\hat{a}$ possesses the kernel property.

For all $S$, every typical model of $S$ includes $tk(S)$.

[**Proof.**]{} Suppose a typical model $m$ of $S$ does not include $tk(S)$ as it contains a negation $\neg \hat{a}$ of a kernel typical atom $\hat{a} \in tk(S)$. Due to the kernel property of $\hat{a}$, $S$ has a model $\mu$ that is $\hat{a}$-neighbor of $m$ and hence contains $\hat{a}$. So $T(m) \subset T(\mu)$. Hence, $m$ is not a typical model — a contradiction. By the same argument every typical model of $S$ includes $tk(S)$. $\Box$

Stable beliefs
--------------

People are in constant quest for knowledge. The available knowledge about the real world is being expanded and deepened. If new knowledge is added to $S$, the set of models of $S$ changes, and so the set of possible worlds $W$ changes as well. Indeed, the 