atom $\mathbf{\hat{a}}$ every model of $S$ containing $\neg \mathbf{\hat{a}}$ has an $\mathbf{a}$-neighbor in $MN(S \cup \{\mathbf{\hat{a}}\})$, that is $$\label{kern-prop}
MOD(S \cup \{\neg \mathbf{\hat{a}}\}) =
MN(S \cup \{\neg \mathbf{\hat{a}}\}),$$ then call $\mathbf{\hat{a}}$ *a kernel atom* possessing *the kernel property* (\[kern-prop\]), and let *the typical kernel* of $S$, $tk(S)$, be the set of all kernel atoms of $S$. Figure 1 illustrates the kernel property. $\Box$

By the kernel property, $tk(S)$ includes all atoms $b$ such that $S \models b$ since $MOD(S \cup \{\neg b\}) = \emptyset$.

Let us say that a formula $\phi$ *cancels* all models of $S$ in which $\phi$ is false.

\[lemma\] For all kernel atoms $\hat{a}$ of $S$ and all literals $l \not= \neg\hat{a}$, if $l$ is consistent with $S$ then $l$ is so with $S \cup \{\hat{a}\}$.

[**Proof.**]{} Suppose $l$ is consistent with $S$, but inconsistent with $S \cup \{\hat{a}\}$, and so cancels all models of $MOD(S \cup \{\hat{a}\})$ including $MN(S \cup \{\hat{a}\})$. Since $l \not= \neg \hat{a}$, $l$ cancels all models of $MN(S \cup \{\neg \hat{a}\})$ as well. By the kernel property of $\hat{a}$, $MN(S \cup \{\neg \hat{a}\}) = MOD(S \cup \{\neg \hat{a}\})$. So $l$ cancels all models of $MOD(S)$ and becomes inconsistent with $S$ — a 