Splitting and Updating Hybrid Knowledge Bases (Extended Version) 23 from the inductive hypothesis for ψ:

$$\langle I,M,N\rangle\models\phi\Longleftrightarrow(\forall J\in M)\left(\langle J,M,N\rangle\models\psi\right)$$ $$\Longleftrightarrow(\forall J\in M)\left(\left\langle J^{[U]},M^{[U]},N^{[U]}\right\rangle\models\psi\right)$$ $$\Longleftrightarrow\left(\forall J\in M^{[U]}\right)\left(\left\langle J,M^{[U]},N^{[U]}\right\rangle\models\psi\right)$$ $$\Longleftrightarrow\left(I^{[U]},M^{[U]},N^{[U]}\right)\models\phi\;\;;$$

6
◦If φ is of the form not ψ, then pr(φ) = pr(ψ), so U ⊇ pr(ψ). The claim follows similarly as in the previous case:

$$\langle I,M,N\rangle\models\phi\Longleftrightarrow(\exists J\in N)\left(\langle J,M,N\rangle\not\models\psi\right)$$ $$\Longleftrightarrow(\exists J\in N)\left(\left\langle J^{[U]},M^{[U]},N^{[U]}\right\rangle\not\models\psi\right)$$ $$\Longleftrightarrow\left(\exists J\in N^{[U]}\right)\left(\left\langle J,M^{[U]},N^{[U]}\right\rangle\not\models\psi\right)$$ $$\Longleftrightarrow\left\langle I^{[U]},M^{[U]},N^{[U]}\right\rangle\models\phi\.\qed$$

Corollary 26 Let T be a set of formulae, U be a set of predicate symbols such that U ⊇ pr(T ) and M, N ∈ M be such that they coincide on U. Then

$$M\models\mathcal{T}\Longleftrightarrow N\models\mathcal{T}$$

Proof We will prove the equivalence only in one direction, the proof of the second direction can be written analogically.

Suppose that M |= T . Then for every φ ∈ T and all I ∈ M we have hI, M, Mi |= φ.

We want to show that N |= T . Let's pick some φ ∈ T and some J ∈ N. Since M[U] =
N[U], there must be some I ∈ M such that I
[U] = J
[U]. By assumption, U contains pr(φ)
and hI, M, Mi |= φ, so Proposition 25 yields

$$\left\langle I^{[U]},M^{[U]},M^{[U]}\right\rangle\models\phi\.$$  As mentioned above, $I^{[U]}=J^{[U]}$ and $M^{[U]}=N^{[U]}$, so  $$\left\langle J^{[U]},N^{[U]},N^{[U]}\right\rangle\models\phi\.$$

Another application of Proposition 25 now yields hJ, N, Ni |= φ and since J and φ were chosen arbitrarily, we can conclude that N |= T .

Proposition 27 Let U be a set of predicate symbols and M, N ∈ M. Then the following implications hold:
(1) If M ⊆ N, then M[U] ⊆ N[U].

(2) If M = N, then M[U] = N[U]. (3) If M ⊆ N and M[U] ( N[U], then M ( N.

Proof