3. If D$_{1}$ is preordered then D$_{2}$ is preordered. Proof. (1) Suppose D$_{1}$ is balanced and wx, xy, yz, wz ∈ A ( D$_{2}$ ) for some w, x, y, z ∈ V ( D$_{2}$ ). Set a = θ ( w ), b = θ ( x ), c = θ ( y ), and d = θ ( z ). Then ab, bc, cd, ad ∈ A ( D$_{1}$ ) by property 1 of Definition 1.1. If xz ∈ A ( D$_{2}$ ) then property 1 of Definition 1.1 gives bd ∈ A ( D$_{1}$ ) so ac ∈ A ( D$_{1}$ ) since D$_{1}$ is balanced. Then ( a, b, c ) ∈ Trans ( D$_{1}$ ) so ( w, x, y ) ∈ Trans ( D$_{2}$ ) by Lemma 1.3. This gives wy ∈ A ( D$_{2}$ ). For the other direction assume wy ∈ A ( D$_{2}$ ). Then property 1 of Definition 1.1 gives ac ∈ A ( D$_{1}$ ) so bd ∈ A ( D$_{1}$ ) since D$_{1}$ is balanced. Then ( b, c, d ) ∈ Trans ( D$_{1}$ ) so ( x, y, z ) ∈ Trans ( D$_{2}$ ) by Lemma 1.3. This gives xz ∈ A ( D$_{2}$ ). (2) Suppose D$_{1}$ is stable and wx, wy, xy, xz, yz ∈ A ( D$_{2}$ ) for some distinct vertices w, x, y, z ∈ V ( D$_{2}$ ). Set a = θ ( w ), b = θ ( x ), c = θ ( y ), and d = θ ( z ). Then ab, ac, bc, bd, cd ∈ A ( D ∗ $_{1}$) by property 3 of Definition 1.1. In particular a, b, c, d are distinct and ad ∈ A ( D ∗ $_{1}$) since D$_{1}$ is stable. Then ( a, c, d ) ∈ Trans ( D$_{1}$ ) so ( w, y, z ) ∈ Trans ( D$_{2}$ ) by Lemma 1.3. This proves wz ∈ A ( D ∗ $_{2}$) as desired. Therefore D$_{2}$ is stable if D$_{1}$ is stable. (3) Part 3 follows immediately from Lemma 1.3. The converse does not hold for every part of Theorem 2.2. Figure 3 shows two directed graphs which are not balanced. However, they are both compressions of preordered directed graphs. The compressions are defined in a similar fashion as Example 1.2 by constructing a directed graph which splits the middle vertex in two. 6