18 **Martin Gebser and Joohyung Lee and Yuliya Lierler**
and

$\lambda$ $Y$) = . 
X ∩ (A \ Y ) = ∅ . **(35)**
From (32) and (35), it follows that

$\mathfrak{l}(Z)$. 
X ∩ (A \ Z) = ∅ . **(36)**
Finally, from (31), (33), (34), and (36), we conclude that X |= ES Π(Z).

Corollaryd 1 For any disjunctive program Π, any set X **of atoms, and any elementarily unfounded**
set Y for Π w.r.t. X, X does not satisfy ES Π(Y ), but satisfies ES Π(Z**) for all**
nonempty proper subsets Z of Y .

Proof. From the definition of an elementarily unfounded set, X 6|= ES Π(Y **), and,**
by Propositiond 5, X |= ES Π(Z**) for all nonempty proper subsets** Z of Y .

Theoremd 2 For any disjunctive program Π and any sets X, Y of atoms, Y **is an elementarily** unfounded set for Π w.r.t. X iff Y **is minimal among the nonempty sets of atoms** occurring in Π that are unfounded by Π w.r.t. X.

Proof**. From left to right follows immediately from Corollary**d 1.

From right to left: Assume that Y **is minimal among the nonempty unfounded** sets for Π w.r.t. X whose atoms occur in Π. If Y **is a singleton, it is elementarily** unfounded by Π w.r.t. X. Otherwise, if Y 6⊆ X, there is an atom a ∈ (Y \ X**), and** one can check that (Y \{a}) is also unfounded by Π w.r.t. X**, which contradicts that** Y is a minimal nonempty unfounded set for Π w.r.t. X**. Hence, from the minimality**
assumption on Y , it follows that Y is a subset of X**. It also holds that** X |= ES Π(Z)
for every nonempty proper subset Z of Y **, so that there is a rule (12) in Π such** that A ∩ Z 6= ∅ , **(37)**

$$B\cap Z=\varnothing\ ,$$
$\left(38\right)$ . 
B ∩ Z = ∅ , **(38)**
$$X\models B,F\ ,$$
$\left(39\right)$ . 
$\left(40\right)^3$

X |= B, F , **(39)**
$$\operatorname{and}$$
$$X\cap(A\setminus Z)=\varnothing\ .$$
X ∩ (A \ Z) = ∅ . **(40)**
From (40), since $Z\subseteq Y\subseteq X$,  . 
$$A\cap(Y\setminus Z)=\varnothing$$
$\left(41\right)^{2}$
A ∩ (Y \ Z) = ∅ **(41)**
and

$$X\cap(A\setminus Y)=\varnothing\ .$$
X ∩ (A \ Y ) = ∅ . **(42)**
Since Y is unfounded by Π w.r.t. X**, from (37), (39), and (42), it follows that**

$\left(42\right)^{\frac{1}{2}}$
B ∩ Y 6= ∅ ,