\frac{2^{|Base(S)|}(1 - 2^{-k})}{|MOD(S)|} \le
E(S, A^{(k)}_{\wedge}) \le 
\min \left( 1, \frac{2^{|Base(S)| - k}}{|MOD(S)|} \right).$$ Expressions (9) and (10) complete the proof. $\Box$

Let $\mathcal{F}^{(k)}$ be a set of $k$ formulas, and $\mathcal{F}^{(k)}_{\wedge} = \bigwedge_{F \in \mathcal{F}^{(k)}} F$. If $A^{(k)}$ is replaced with $\mathcal{F}^{(k)}$ then Theorem 4.1 implies the following

\[corollary\] (i) $E(S, \mathcal{F}^{k}_{\wedge}) \ge 
\sum_{F \in \mathcal{F}^{(k)}} E(S, F) - k + 1$;

\(ii) For all formulas $\phi, \psi$, if $E(S, \hat{\phi}) > 0.5$ then $\hat{\phi} \wedge
\hat{\psi}$\
is consistent with $S$;

\(iii) If $E(S, F) = 0.5$ call $F$ a *neutral* formula. If there are two neutral formulas in a set $\mathcal{F}^{(k)}$ then $\mathcal{F}^{(k)}_{\wedge}$ may be inconsistent with $S$. $\Box$

Typical models
--------------

Let $T(S)$ denote the set of all typical atoms of $S$, and $T(m)$ be the set of all typical atoms contained in a model $m$: $$T(S) = \{\hat{a} | a \in Base(S)\}, \quad 
T(m) = \{\hat{a} | \hat{a} \in m\} = T(S) \cap m.$$

\[def-typmod\] If there exists a model $\mu$ of $S$ such that $T(\mu) = T(S)$ then $\mu$ is *the most typical model* of $S$. For all $m \in MOD(S)$, if there is no model $m'$ of $S$ such that $T(m) \subset T(m')$ then 