\alpha = \sum_{l \in A^{(k)}} E(S, l) & - & k + 1,\\
\beta = 1 - \frac{2^{|Base(S)| - k}(2^k - 1)}{|MOD(S)|},
\quad \gamma & = & \frac{2^{|Base(S)| - k}}{|MOD(S)|}.\end{aligned}$$

[**Proof**]{}. First, we prove by induction on $k$ that $$|MOD(S \cup A^{(k)})| \ge \sum_{l \in A^{(k)}} |MOD(S \cup
 \{l\})| - (k - 1)|MOD(S)|.$$

*Base*. For $k = 1$ inequality (8) holds trivially.

*Step*. Let $M_1, M_2$ be subsets of $MOD(S)$, then $$|M_1 \cap M_2| \ge |M_1| + |M_2| - |MOD(S)|. \nonumber$$ If inequality (8) holds for all $1 \le i \le j$, then it holds for $j + 1$. Indeed, let $A^{(j+1)} = A^{(j)} \cup \{l'\}$, then $$|MOD(S \cup A^{(j+1)})| = |MOD(S \cup A^{(j)}) \cap MOD(S 
\cup \{l'\})| \ge \nonumber$$ $$\sum_{l \in A^{(j)}} |MOD(S \cup \{l\})| -
(j - 1)|MOD(S)| + |MOD(S \cup \{l'\})| - |MOD(S)| = \nonumber$$ $$\sum_{l \in A^{(j + 1)}} |MOD(S \cup \{l\})| - j|MOD(S)|. \nonumber$$ Further, by (8) and since evidence is a non-negative value, $$E(S, A^{(k)}_{\wedge}) = \frac{|MOD(S \cup A^{(k)})|}{|MOD(S)|} 
\ge \max \left( 0, \sum_{l \in A^{(k)}} E(S, l) - k + 1 \right).$$

Next, $A^{(k)}_{\wedge}$ is true in $2^{|Base(S)| - k}$ interpretations of $S$ but false in the rest of them. So $$1 - 