all $m \in MOD(S)$\
$ER(\mu) \le ER(m)$.

[**Proof.**]{} (i) If $m$ is a typical model of $S$ then (i) holds trivially, else there is a typical model $\mu$ such that $T(m) \subset
T(\mu)$. Denote $\delta_1 = T(\mu) - T(m) = \mu - m$, $\delta_2 = m - \mu$, $B = |Base(S)|$. All literals of $\delta_1$ are typical atoms. There is a bijection between $\delta_1$ and $\delta_2$ such that to every literal $\hat{a} \in \delta_1$ corresponds $\neg \hat{a} \in \delta_2$. Since for all $a \in Base(S)$ $E(S, \hat{a}) \ge E(S, \neg\hat{a})$, $$ER(\mu) - ER(m) = \frac{1}{B} 
\sum_{\hat{a} \in \delta_1} (E(S, \neg\hat{a}) -
E(S, \hat{a})) \le 0.$$

\(ii) If $\mu$ is the most typical model of $S$ then for all $m \in MOD(S)$\
$T(m) \subset T(\mu)$, hence $ER(\mu) \le ER(m)$. $\Box$

By Theorem 4.2, if there exists the most typical model of $S$ then it is the most trustworthy one among all models of $S$. Otherwise there is a typical model with a minimum value of erratum among all models of $S$.

Let $ER(mtm)$, $ER(rand)$, $ER(worst)$, $E(S)$ denote respectively the erratum of the most typical model, the expected erratum of a randomly chosen model, the erratum of a model containing no typical 