∪ GUSM ∪N,Π∪ch(N)\N−

$N\cup(M\cup\{a\mid a\gets B\in\Pi\cup ch(N)\text{and}B\subseteq M\cup N\}\setminus N)$  $\cup(\overline{GUS(M\cup N,\Pi\cup ch(N))}\setminus\{\neg a\mid a\in N^{-}\})$  $N\cup M\cup\{a\mid a\gets B\in\Pi\cup ch(N)\text{and}B\subseteq M\cup N\}$  $\cup\overline{GUS(M\cup N,\Pi\cup ch(N))}$.  
Since this last set is consistent, it is equal to WΠ∪ch(N)(M ∪N) = WΠ∪ch(N)(WiΠ(N)
(/0)∪N).

Applying the induction hypothesis, the inductive step follows.

Proposition 2 For a PC(ID) theory (F,Π) such that Π *is a normal program, M is a model* of (F,Π) if and only if M is a model of (F,Π) *according to the definition in (Denecker* 2000).

Proof Let (F,Π) be a PC(ID) theory. In (Denecker 2000), the authors state that a consistent and complete (over At(F ∪Π)) set M of literals is a model of (F,Π) if
(i) M is a model of F, and
(ii) M = W
fix Π(MOΠ)
(/0)∪ MOΠ.

To prove the assertion it is sufficient to show that for any model M of F such that |M| =
At(Π ∪F), M = W
fix Πo (MOΠ) if and only if M = W
fix Π(MOΠ)
(/0)∪ MOΠ. Let N = MOΠ. The definitions of O
Π and Πo directly imply that |N|∩*Head*(Π) = /0 and that Πo = Π∪ch(N).

Thus, the property follows from Lemma 3.

## 8.2 Proofs Of Results From Section 3

Proposition 3 For a logic program Π *and a set X of atoms,*
(a) X ⊆ Head(Π) and X is an input answer set of Π *if and only if X is an answer set of* Π.

(b) if (X \ *Head*(Π)) ∩ At(Π) = /0, then X is an input answer set of Π if and only if X ∩Head(Π) *is an answer set of* Π.

Proof The proof of part (a) is straightforward and follows directly from the definition of an input answer set. To prove (b), let us assume first that X is an input answer set of Π. By the definition, X is an answer set of Π∪(X\*Head*(Π)). Thus, X is the least model of the reduct
[Π∪(X\*Head*(Π))]X. Clearly, we have [Π∪(X\*Head*(Π))]X = ΠX∪(X\*Head*(Π)). Since
(X \ *Head*(Π)) ∩ At(Π) = /0, ΠX = ΠX∩*Head*(Π). It follows that X is the least model of ΠX∩Head(Π)∪(X \*Head*(Π)). Using again the assumption (X \*Head*(Π))∩At(Π) = /0, one can show that X ∩ *Head*(Π) is the least model of ΠX∩*Head*(Π). Thus, X ∩ *Head*(Π) is an answer set of Π
The proof in the other direction is similar. Let us assume that X ∩ *Head*(Π) is an answer set of Π. It follows that X ∩ *Head*(Π) is the least model of ΠX∩*Head*(Π). Since (X \
Head(Π))∩ At(Π) = /0, X is the least model of ΠX∩Head(Π) ∪ (X \ *Head*(Π)). Moreover, since ΠX∩*Head*(Π) = ΠX, X is the least model of ΠX∪(X\*Head*(Π)) = [Π∪(X\*Head*(Π))]X.

Thus, X is an input answer set of Π.