By Proposition 34 we know that M ⊆ Y and that Y is saturated relative to P \ U. We need to show that Y is an MKNF model of eU (K, X). By Proposition 51, this holds if and only if Y is and S5 model of π(eU (K, X)) and for every Y
′ ) Y it holds that hY
′, Y **i 6|**=
π(eU (K, X)). The former follows directly from Corollary 61, so we will focus on the latter condition.

Let's pick some Y
′ ) Y . By Proposition 44 there exists the greatest MKNF interpretation M′that coincides with M on U (i.e. M′[U] = M[U]) and with Y
′ on P \ U (i.e.

M′[P\U] = Y
′[P\U]) and which contains M ∩ Y
′. Hence,

$$\gamma^{\prime}\subseteq M^{\prime}$$

M ⊆ M ∩ Y ⊆ M ∩ Y
′ ⊆ M′. (B6)
Furthermore, we know that Y is saturated relative to P \ U, so we can use Proposition 30(2) to conclude that M[P\U] = Y
′[P\U] ( Y
′[P\U] = M′[P\U]. (B7)
Consequently, by (B6), (B7) and Proposition 27(3), we obtain M ( M′. This, together with the assumption that M is an MKNF model of K, implies that hM′, M**i 6|**= π(K). We can now apply Corollary 63 to conclude that hY
′, Y i 6|= π(eU (K, X)), which is also the desired conclusion.

Proposition 68 Let U be a splitting set for a hybrid knowledge base K and h**X, Y** i be a solution to K with respect to U. Then X ∩ Y is an MKNF model of K.

Proof Let K = hO,Pi and M = X ∩ Y . In order to show that M is an MKNF model of K, we need to prove that M |= π(K) and that for every M′ ) M it holds that hM′, M**i 6|**= π(K).

We will verify the two conditions separately.

Since h**X, Y** i is a solution to K with respect to U, X must an MKNF model of bU (K)
and Y an MKNF model of eU (K, X). So X |= π(bU (K)) and Y |= π(eU (K, X)). Consequently, by Corollary 61, M |= π(K).

We know that X is an MKNF model of bU (K), so, by Proposition 29, X is saturated relative to U. Similarly, since Y is an MKNF model of eU (K, X), it must be saturated relative to P \ U. Hence, by Proposition 40, M is semi-saturated relative to U, M[U] =
X[U]and M[P\U] = Y
[P\U].

Now take some MKNF interpretation M′ ) M and let X′ = X ∪ M′and Y
′ =
Y ∪ M′. We already inferred that M is semi-saturated relative to U, which means that by Proposition 41(2) one of the following cases must occur:
a) If M′[U] ) M[U], then X′[U] = X[U] ∪ M′[U] = M′[U] ) M[U] = X[U],
and so Proposition 27(3) implies that X′ ) X. Hence, since X is an MKNF model of bU (K), we infer that hX′, X**i 6|**= π(bU (K)) and by Corollary 64 we obtain hM′, M**i 6|**= π(K), which is what we wanted to prove.

b) If M′[U] = M[U]and M′[P\U] ) M[P\U], then X′[U] = M′[U] = M[U]and Y
′[P\U] = Y
[P\U] ∪ M′[P\U] = M′[P\U] ) M[P\U] = Y
[P\U],