\hat{a}_i \}) = 
MOD(S \cup \{\neg \hat{a}_i \}) - MOD(S \cup \{\neg \hat{a}_1 \}) =
MOD(S_1 \cup \{\neg \hat{a}_i \})$.\
Hence $\hat{a}_i$ has the kernel property in $S_1$.

*Step*. Suppose kernel atoms $\hat{a}_1, \ldots , \hat{a}_i$ ($1 \le i < k$) have been included in $S$ such that $S_i = S \cup \{\hat{a}_1, \ldots , \hat{a}_i\}$. Then by the same argument as above $S_i$ is consistent, and for all $i < j \le k$ we have $\hat{a}_j \in tk(S_i)$. Hence $S_k = S \cup tk(S)$ is consistent.

So all kernel atoms of $S$ can be included into $S$ in any order preserving consistency of the augmented set. However, this may not be true regarding a non-kernel typical atom $\hat{b}$ of $S$ such that $\hat{b} \not\in tk(S)$. Since $b$ does not possess the kernel property, $MN(S \cup \{\neg \hat{b}\}) \subset MOD(S \cup \{\neg \hat{b}\})$. So unlike the situation described by Lemma 4.1, inclusion into $S$ of $tk(S)$ (or even of any literal $l$ consistent with $S$) may cancel all models of $MOD(S \cup \{\hat{b}\})$ and of $MN(S \cup \{\neg \hat{b}\})$. Since the latter set is just a proper subset of $MOD(S \cup \{\neg \hat{b}\})$, we get $MOD(S \cup tk(S)) = MOD(S 