atoms from Q in the scope of negation, and since all Y
′ whose intersection generate Z are superset of Z with Y
′ \ Z ⊆ Q, all those Y
′satisfy Def(q)
X by Lemma 9.

By property (ii), we have that q ∈ Y
′for all Y
′, and then q ∈ Z.

It remains to prove claim (c). Take any q ∈ Z that belongs to Q ∩ X. Set Y
′ = Z \ {q} satisfies condition (i), but it cannot satisfy (ii), because sets Y
′
that satisfy (i) and (ii) are supersets of Z by construction of Z. Consequently, Y
′ 6|= Def(q)
X. Since all positive occurrences of atom q in Def(q) are in the scope of negation and Y
′ = Z \ {q}, we can conclude that Z 6|= Def(q)
X by Lemma 9 again.

Now consider two cases. If X 6|= Γ2 then clearly X is not a stable model of Γ2. It is not a stable model of Γ1 as well. Indeed, since X |= Γ1, we have that, for some q ∈ Q ∩ X, X 6|= Def(q). Consequently, Def(q)
X = ⊥ and then X 6|= ∆,
but, since Z |= ∆ by (c) and Z ⊆ Y ⊆ X by (a), Z is a proper subset of X. Since Z |= ΓX
1 by (b), X is not a stable model of Γ1.

In the other case (X |= Γ2) it is not hard to see that Γ
X 2 is equivalent to Γ
X
1 ∪∆.

We have that Z |= ΓX
1by (b), and then Z |= ΓX
2by (c). Since Y 6|= ΓX
2
, Z 6= Y .

On the other hand, Z ⊆ Y ⊆ X by (a). This means that Z is a proper subset of X that satisfies Γ
X
1 and Γ
X
2
, and we can conclude that X is not an stable model of any of Γ1 and Γ2.

## 4.6 Proof Of Proposition 11 Lemma 1. Rule

l1 ∧ · · · ∧ lm → a1 **∨ · · · ∨** an (37)
(n > 0, m ≥ 0*) where* a1, . . . , an *are atoms and* l1, . . . , lm are literals, is strongly equivalent to the set of n *implications* (i = 1, . . . , n)

$$i))\to a_{i}.$$

(l1 ∧ · · · ∧ lm ∧ (a1 → ai) **∧ · · · ∧** (an → ai)) → ai. (38)
Proof. Let F be (37) and Gi (i = 1, . . . , n) be (38). We want to prove that F is strongly equivalent to {G1, . . . , Gn} by showing that F
X is classically equivalent to {GX
1
, . . . , GX
n
}. Let H be l1 **∧ · · · ∧** lm.

Case 1: X 6|= H. Then the antecedents of F and of all Gi are not satisfied by X. It is then easy to verify that the reducts of F and of all Gi relative to X
are equivalent to ⊤. **Case 2**: X |= H and X 6|= F. Then clearly F
X = ⊥. But, for each i, GX
iis ⊥: indeed, since X 6|= F, X 6|= ai for all i = 1, . . . , n. It follows that the consequent of each Giis not satisfied by X, but the antecedent is satisfied, because X |= H and in each implication aj → aiin Gi, the antecedent

$$(38)$$
