(b) for any set X of atoms, GX *is classically equivalent to* AX.

Proof. Part (a) is immediate from Lemma 10, as G = Aˆ. For part (b), we need to show that AˆX is classical equivalent to AX. By Lemma 11, AˆX is classically equivalent to AˆX . It remains to notice that AˆX is classically equivalent to AX by Lemma 10.

Lemma 12. *For any aggregate* oph{F1 = w1, . . . , Fn = wn**}i ≺** N*, formula (9)*
is classically equivalent to

$$\bigwedge_{I\subseteq\{1,...,n\}\,:\,o p(\{w_{i}:i\in I\})\neq N}\left(\bigvee_{i\in\overline{{I}}}F_{i}\right)$$
(40)  $$\begin{array}{l}\mbox{\rm(40)}\end{array}$$ . 
if the aggregate is monotone, and to

$$\bigwedge_{I\subseteq\{1,...,n\}:o p(\{w_{i}:i\in I\})\neq N}(\neg\bigwedge_{i\in I}F_{i})$$

if the aggregate is antimonotone.

Proof. Consider the case of a monotone aggregate first. Let G be (9), and H
be (40). It is easy to verify that H entails G. The opposite direction remains.

Assume G, and we want to derive every conjunctive term

$$\bigvee_{i\in I}F_{i}$$
$$(41)$$
Fi (41)
in H. For every conjunctive term D of the form (41) in H, op({wi: i ∈ I}) 6≺ N.

As the aggregate is monotone then, for every subset I
′ of I, op({wi: i ∈ I
′}) 6≺
N, so that the implication

$$(\bigwedge_{i\in I^{\prime}}F_{i})\to(\bigvee_{i\in\overline{{{I^{\prime}}}}}F_{i})$$

is a conjunctive term of H for all I
′ ⊆ I. Then, since I

$$:\overline{{{I^{\prime}}}}=\overline{{{I}}}\cup(I\setminus I^{\prime}),$ ("$\Rightarrow$")
denotes entailment, and "⇔" equivalence)