contradiction. $\Box$

(110,80)

(15,20)[(40,15)[$MN(S \cup \{\neg\hat{a}\})$]{}]{} (55,20)[(40,15)[$MN(S \cup \{\hat{a}\})$]{}]{} (55,35)[(0,1)[20]{}]{} (95,35)[(0,1)[20]{}]{} (55,55)[(1,0)[40]{}]{} (40,7.5)[(-2,1)[23]{}]{} (70,7.5)[(2,1)[23]{}]{} (30,45)[(-3,-2)[13]{}]{} (40,45)[(3,-2)[13]{}]{} (70,65)[(-3,-2)[13]{}]{} (80,65)[(3,-2)[13]{}]{} (47,5)[$MOD(S)$]{} (20.5,47.5)[$MOD(S \cup \{\neg\hat{a}\})$]{} (61.5,67.5)[$MOD(S \cup \{\hat{a}\})$]{}

\[th-kernel\] For all $S$, $tk(S)$ is consistent with $S$. There is no superset of $tk(S)$ possessing this property.

[**Proof.**]{} By induction on the serial number of kernel atoms of $S$ numbered arbitrarily in $tk(S) = \{\hat{a}_1, \ldots , \hat{a}_k\}$.

*Base*. Include $\hat{a}_1$ into $S$ producing $S_1 = S \cup \{\hat{a}_1\}$; $\hat{a}_1$ is consistent with $S$, so $S_1$ is consistent; but $\hat{a}_1$ cancels all models of $S$ containing $\neg \hat{a}_1$ such that

$MOD(S_1) = MOD(S) - MOD(S \cup \{\neg \hat{a}_1\}) 
= MOD(S \cup \{\hat{a}_1\}) \not= \emptyset,$\
$MOD(S_1 \cup \{\hat{a}_1\}) = MOD(S_1), \quad \quad
MOD(S_1 \cup \{\neg \hat{a}_1\}) = \emptyset.$

It turns out that for $1 < i \le k$ every $\hat{a}_i \in tk(S)$ is a kernel atom of $S_1$. Indeed,

\(i) by Lemma 4.1 $\hat{a}_i$ is consistent with $S_1$ since it is consistent with $S$;\
(ii)

$MN(S_1 \cup \{\neg \hat{a}_i\}) =
MN(S \cup \{\neg \hat{a}_i\}) - MOD(S \cup \{\neg \hat{a}_1\})$;

\(iii) since $\hat{a}_i$ is a kernel atom of $S$,

$MN(S \cup \{\neg \hat{a}_i\}) = MOD(S \cup \{\neg \hat{a}_i\})$.

So (i) - (iii) imply\
$MN(S_1 \cup \{\neg 