Proof. First, we prove by induction on k **that**

$$MOD(S\cup A^{(k)})|\geq\sum_{l\in A^{(k)}}|MOD(S\cup\{l\})|-(k-1)|MOD(S)|.\tag{8}$$

Base. For k **= 1 inequality (8) holds trivially.**
Step. Let M1, M2 be subsets of MOD(S**), then**

$=\;\left[10-1\right]$ . 
|M1 ∩ M2| ≥ |M1| + |M2| − |MOD(S)|.

If inequality (8) holds for all 1 ≤ i ≤ j, then it holds for j **+ 1. Indeed, let**
A(j+1) = A(j) ∪ {l
′}**, then**
|MOD(S ∪ A
(j+1))| = |MOD(S ∪ A
(j)) ∩ MOD(S ∪ {l
′})| ≥

$$\sum_{l\in A^{(j)}}|M O D$$

|MOD(S ∪ {l})| − (j − 1)|MOD(S)| + |MOD(S ∪ {l
′})| − |MOD(S)| =

$\lambda_{1}=-\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$

$$\sum_{l\in A^{(j+1)}}|M O D(S\cup\{l\})|-j|M O D(S)|.$$
Further, by (8) and since evidence is a non-negative value,

$$E(S,A_{\wedge}^{(k)})=\frac{|MOD(S\cup A^{(k)})|}{|MOD(S)|}\geq\max\left(0,\sum_{l\in A^{(k)}}E(S,l)-k+1\right).\tag{9}$$

Next, A
(k)
∧ is true in 2|Base(S)|−kinterpretations of S **but false in the rest of**
them. So

$$1-\frac{2^{|Base(S)|}(1-2^{-k})}{|MOD(S)|}\leq E(S,A_{\wedge}^{(k)})\leq\min\left(1,\frac{2^{|Base(S)|-k}}{|MOD(S)|}\right).\tag{10}$$

Expressions (9) and (10) complete the proof. ✷
Let F
(k) be a set of k **formulas, and** F
(k)
∧ =VF ∈F(k) F. If A(k)**is replaced**
with F
(k)**then Theorem 4.1 implies the following**
Corollary 4.1 (i) E(S, F
k∧) ≥PF ∈F(k) E(S, F) − k + 1;
(ii) For all formulas φ, ψ, if E(S, φˆ) > 0.5 **then** φˆ ∧ ψˆ
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 F
(k)**then** F
(k)
∧ **may be inconsistent with** S. ✷

## 4.2 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) = {aˆ|a ∈ Base(S)}, T(m) = {aˆ|aˆ ∈ m} = T(S) ∩ m. **(11)**