Intuitively, a ranking structure of X is just a partitioning P of X equipped with a total ordering which orders the partitions in P.

Proposition 6. Any stratified order ✁ *on a set X can be uniquely determined by a* ranking structure *of X.*
Proof. Similarly to the ideas from Proposition 4 and Proposition 5, we define an equivalence relation from the stratified order ✁ as follows:
≃✁
df= {(x,y) ∈ X ×X : ¬x✁y∧ ¬y✁x∧x 6= y}
Then let P be the set of all partitions of X with respect to this equivalence relation ≃✁.

Next we define the relation ◭ as ◭
df= {(A,B) ∈ P×P : A×B ⊆ ✁}. Then, similarly to Proposition 5, we can check that ◭ is a total ordering.

To recover the stratified order ✁ from the ranking structure (P,◭), we simply reconstruct
✁ =
[{X ×Y : X 6= Y ∧X ◭ Y}.

For a set A, we let K(A) denote the complete graph induced by A. In other words, K(A) = (A,E) and E
df= {(u,v) ∈ A×A : u 6= v}.

For each ranking structure R = (P,◭) of a set X, we have two kinds of graph associated with it:
G(R) = (X,E), where E =
[{X ×Y : X 6= Y ∧X ◭ Y}
Gb(R) = G(R)∪
[
A∈P
K(A)
Intuitively, the graph G(R) is simply the transitive graph of the stratified order encoded by R. And the graph SA∈PK(A) is exactly the graph IC(G(R)), but in this case it is more intuitive to characterize it as the union of complete graphs.

Putting everything together we have the following characterization of the class of all models of Tgso.

Definition 5. Let Mgso denote the class of all possible models for Tgso*. Then any model* M ∈ Mgso *is uniquely determined from*
- *the sets E, EO, and O*
- the graphs G1, G2 *and G*3
- *a family F of ranking structures on EO indexed by the set O, i.e., F* = {Ro : o ∈ O},
such that 1. all conditions from Definition 1 are satisfied
⊓⊔