26 *M. Slota and J. Leite and T. Swift* We will prove that 1. implies 2., 2. implies 3. and finally that 3. implies 1.

Suppose N = σ(**M, U**). Then N
[U] = { I
[U]| I **∈ I ∧** I
[U] ∈ M[U]} = M[U],
so N coincides with M on U. Furthermore, any I ∈ I with I
[U] ∈ N[U] must also satisfy I
[U] ∈ M[U]. Thus, I ∈ N, so N is saturated relative to U. This shows that 1. implies 2.

To show that 2. implies 3., suppose N coincides with M on U and is saturated relative to U. Suppose N′ ∈ M coincides with M on U and I ∈ N′. Then I
[U] ∈ N
′[U] = M[U] = N
[U],
so, since N is saturated relative to U, we can conclude that I belongs to N. Consequently, N′is contained in N, so N is the greatest among all N′ ∈ M coinciding with M on U.

Finally, suppose N is the greatest among all N′ ∈ M coinciding with M on U. It can be easily seen that σ(**M, U**) coincides with M on U, so σ(**M, U**) must be a subset of N. It remains to show that N is a subset of σ(**M, U**). But that is an easy consequence of the fact that for any I ∈ N, I
[U] must belong to M[U].

It still remains to show that M is a subset of N if N satisfies one of the above conditions.

We already know that the conditions are equivalent, so we only need to consider one of them. So suppose N = σ(**M, U**) (condition 1.). It can be easily seen from the definition of σ(**M, U**) that every I ∈ M belongs also to N. Hence, M is a subset of N.

Proposition 35 Let U1, U2 be sets of predicate symbols and M ∈ M. Then σ(σ(M, U1), U2) = σ(**M, U**1 ∩ U2) .

Proof Consider the following sequence of equivalences:
I ∈ σ(σ(M, U1), U2) ⇐⇒ I
[U2] ∈ σ(**M, U**1)
[U2]

$\sigma_{\mu\nu}\sim\sigma_{\mu\nu}$ (2.1)

$\Longleftrightarrow(\exists J\in\sigma(M,U_{1}))(J^{[U_{2}]}=I^{[U_{2}]})$  $\Longleftrightarrow(\exists J\in\mathcal{I})((\exists K\in M)(K^{[U_{1}]}=J^{[U_{1}]}\wedge J^{[U_{2}]}=I^{[U_{2}]})$  $\Longleftrightarrow(\exists K\in M)(\exists J\in\mathcal{I})(J^{[U_{1}]}=K^{[U_{1}]}\wedge J^{[U_{2}]}=I^{[U_{2}]})$
Moreover, we also obtain the following:
I ∈ σ(**M, U**1 ∩ U2) ⇐⇒ (∃K ∈ M)(K[U1∩U2] = I
[U1∩U2]) .

So it remains to show that
(∃J ∈ I)(J
[U1] = K[U1] ∧ J
[U2] = I
[U2])
holds if and only if

K[U1∩U2] = I
[U1∩U2].