Proof. Suppose that S is a rule-representable set of SE-interpretations. Then there exists some rule r such that S = modSE (r). Let the sets of interpretations L1, L2 be defined as follows:

$L_{1}=\left\{\ I\in\mathcal{I}\ |\ B(r)^{+}\subseteq I\subseteq\mathcal{L}\setminus H(r)^{+}\ \right\}$  $L_{2}=\left\{\ J\in\mathcal{I}\ |\ H(r)^{-}\cup B(r)^{+}\subseteq J\subseteq\mathcal{L}\setminus B(r)^{-}\ \right\}$
It can be straightforwardly verified that these sets are convex sublattices of hI, ⊆i.

It remains to prove that the complement of S relative to I
SE is equal to the set of SE-interpretations

$$\left\{\,\langle I,J\rangle\in{\mathcal{I}}^{\mathsf{SE}}\,\big|\,I\in L_{1}\wedge J\in L_{2}\,\right\}\cup\left\{\,\langle I,J\rangle\in{\mathcal{I}}^{\mathsf{SE}}\,\big|\,J\in L_{1}\cap L_{2}\,\right\}\ .\quad0$$

According to Lemma 8, an SE-interpretation h**I, J**i does not belong to S if and only if these two conditions are satisfied:
1. H(r)
− ∪ B(r)
+ ⊆ J and J **⊆ L \** B(r)
−.

2. Either J **⊆ L \** H(r)
+ or both B(r)
+ ⊆ I and I **⊆ L \** H(r)
+.

It is not difficult to see that whenever the first condition and first disjunct of the second condition are satisfied, h**I, J**i belongs to the second part of the set (6). Similarly, the first condition together with the second disjunct of the second condition imply that h**I, J**i belongs to the first part of the set (6). Conversely, given the definitions of L1 and L2, it is easy to see that any SE-interpretation belonging to the set (6) satisfies the conditions of Lemma 8. Thus, the set (6) coincides with the complement of S relative to I
SE
.

Now suppose that L1, L2 are two convex sublattices of hI, ⊆i such that the complement of S relative to I
SE is equal to the set (6). Let ⊤1, ⊥1 be the top and bottom elements of L1 and ⊤2, ⊥2 be the top and bottom elements of L2. Furthermore, let r be a rule of the form H+; ∼H− ← B
+, ∼B
−.

where H+ = **L \ ⊤**1, H− = ⊥2, B+ = ⊥1 and B− = **L \ ⊤**2. We will show that S = modSE (r).

Suppose first that the SE-interpretation h**I, J**i is not an SE-model of r. Then, by Lemma 8, J includes H−∪B+ = ⊥2∪⊥1 and J is included in L\B− = L\(L\⊤2) =
⊤2. By convexity of L2 we now obtain that J belongs to L2. Lemma 8 also implies that either J is included in L \ H+ = ⊤1, or I includes B+ = ⊥1 and is included in L \ H+ = ⊤1. The convexity of L1 now implies that in the former case J belongs to L1, while in the latter case I belongs L1. In any of these cases, h**I, J**i is a member of the set (6).

Now let h**I, J**i be some SE-interpretation not belonging to S. If h**I, J**i belongs to the first part of the set (6), then ⊥1 ⊆ I ⊆ ⊤1 and ⊥2 ⊆ J ⊆ ⊤2. Thus, I includes B+ and is included in L \ H+, and J includes H− and is included in L \ B−. Also, since I is a subset of J, J includes B+. Lemma 8 then implies that h**I, J**i is not an SE-model of r. If h**I, J**i belongs to the second part of the set (6), then ⊥1 ⊆ J ⊆ ⊤1 and ⊥1 ⊆ J ⊆ ⊤2. Thus, J includes both H− and B+ and is included in L \ B− and in L \ H+. As a consequence of Lemma 8, h**I, J**i is not an SE-model of r. ⊓⊔