then M is also a model of H(r) ⊂ B+(r), so M is again a model of r**. Consequently,**
M is an S5 model of K. Now take some M′ ) M. Then since M **is the greatest** model of KM , there is some rule rM = (H(r) ⊂ B+(r)) ∈ PM **such that** M′ 6|= rM ,
i.e.

M |= B
−(r) ∧ M′|= B
+(r) ∧ M′6|= H(r)
For any I
′ ∈ M′**, this is equivalent to**

$\sum_{n=1}^{\infty}\left(\frac{1}{n}\right)\left(\frac{1}{n^{2}}\right)\left(\frac{1}{n^{2}}\right)$
$\ell=\mathbb{R}$. 
hI
′, M′, Mi |= B
−(r) ∧ hI
′, M′, Mi |= B
+(r) ∧ hI
′, M′, M**i 6|**= H(r)
which in turn is equivalent to hI
′, M′, Mi 6|= r. So M **is indeed an MKNF model of**
K.

## Appendix D Properties Of The Hybrid Update Operator

Proposition 3.2. Let P be a finite ground definite program, T **a TBox and** A a sequence of ABoxes. Then TP⊕T A **is a continuous function on the complete partial**
order of all subsets of I **with the least element** I. Proof of Proposition 3.2 Consider some directed subset D of M. To prove that TP⊕T A **is continuous, we**
need to show that sup(TP⊕T A(D)) = TP⊕T A**(sup(**D)) .

To simplify notation in this proof, we define for any set of first-order interpretations M **the following set:**

$\mbox{con}(M)=\{\ H^{*}(r)\ |\ r\in{\cal P}\wedge M\models B(r)\ \}$.  
Notice that if M ⊇ N (or M ≤ N **using the partial order on sets of first-order**
intepretations), then con(M) ⊆ con(N).

By definition we now have

$$T_{\mathcal{P}\oplus\mathcal{T}\mathcal{A}}(\sup(\mathcal{D}))=\text{mod}\left(\text{con}\left(\bigcap\mathcal{D}\right)\oplus^{\mathcal{T}}\mathcal{A}\right)$$ (D1) $$=\text{incorporate}^{\mathcal{T}}\left(\mathcal{A},\text{mod}\left(\mathcal{T}\cup\text{con}\left(\bigcap\mathcal{D}\right)\right)\right)$$ $$=\text{incorporate}^{\mathcal{T}}\left(\mathcal{A},\text{mod}(\mathcal{T})\cap\text{mod}\left(\text{con}\left(\bigcap\mathcal{D}\right)\right)\right)$$
$$\operatorname{and}$$
$$\sup(T_{\mathcal{P}\oplus\mathcal{T}\mathcal{A}}(\mathcal{D}))=\bigcap_{M\in\mathcal{D}}\,\mathsf{mod}\left(\mathsf{con}(M)\oplus^{\mathcal{T}}\mathcal{A}\right)$$ $$=\bigcap_{M\in\mathcal{D}}\,\mathsf{incorporate}^{\mathcal{T}}\left(\mathcal{A},\mathsf{mod}(\mathcal{T}\cup\mathsf{con}(M))\right)$$ (D2) $$=\bigcap_{M\in\mathcal{D}}\,\mathsf{incorporate}^{\mathcal{T}}\left(\mathcal{A},\mathsf{mod}(\mathcal{T})\cap\mathsf{mod}(\mathsf{con}(M))\right)$$

First suppose that a first-order interpretation I is in TP⊕T A(sup(D**)). Then by the**
previous equation we have that there is some J ∈ mod(T ) ∩ mod(con(TD)) such