Doubt regarding the term "equivalence" in a research paper (graph theory) In Peter Alles, The dimension of sums of graphs, Discrete Math. 54 (1985), 229–233 (pdf), the author quotes the following proposition:

The dimension of a graph $G$, is the least number of equivalences
$E_1,\dots,E_n$ such that
(1) $E(G^c)=\cup_{i=1}^n E_i$,
(2) $\cap_{i=1}^n E_i=\Delta$
where $G^c$ denotes the complement of the graph $G$, and $\Delta$ the
  diagonal of the relation $V(G)\times V(G)$.

The he makes the following definition:

Let $A=(a_{ij}), B=(b_{ij})$ be $n\times r$ matrices, $n\le r$, and
  $1\le a_{ij},b_{ij}\le n$ for all $1\le i\le n, 1\le j\le r$. $A$ and
  $B$ are said to have covering property if for each
  $k,l\in\{1,\dots,n\}$ there exists $i,j$ such that
  $(k,l)=(a_{ij},b_{ij})$. Further we assume that $A,B$ are column latin (by which I presume that all $n$ numbers occur in each column).

Now consider the graph $2K_n:=K_n^1+K_n^2$ where by $+$ we mean disjoint union of $K_n^1$ with $K_n^2$ which are both complete graphs on $\{1,\dots,n\}$.
My question is as follows:
The author makes the statements:

For $2K_n = K_n^1+K_n^2$ where $V(K_n^i) = \{1,\dots,n\}$, we get a
  partition of the vertex set of $2K_n$, into $r$ equivalences according
  to the Proposition if we apply two $n \times r$ matrices $A, B$ with
  covering property by defining:
$$E_j=\bigcup_{i=1}^{n}\langle a_{ij}, b_{ij}\rangle; j=1,\dots,r$$
since the covering property ensures that each pair of non-adjacent
  vertices of $2K_n$, i.e. one vertex of $K_n^1$ and one vertex of
  $K_n^2$, is contained in at least one class $\langle a_{ij},b_{ij}\rangle$ of an equivalence $E_j$ where the a's belong to $K_n^1$
  and the b's to $K_n^2$. On the other hand, the meet of all
  equivalences is the diagonal of the relation $V(2K_n)\times V(2K_n)$.

Now firstly I cannot understand what is meant by the term equivalence. I believe it must mean an equivalence relation but it is not clear to me that this is an equivalence relation on which set (what is $V(2K_n)$? Is it $\{1,\dots,n\}$?). Secondly exactly how is $E_j$ defined? Thirdly, why do the $E_j$ meet the criteria laid down in the proposition.
 A: Equivalence does indeed mean "equivalence relation". In this case, $2K_n$ is the disjoint union of two complete graphs. $V(2K_n)$ should have $2n$ elements.
Here's the mildly annoying part. The paper doesn't give the $2n$ elements distinct names. Rather, the vertices of one $K_n$ are named $\{1,2,\dots,n\}$, and the vertices of the other $K_n$ are named a different $\{1,2,\dots,n\}$. 
In the definition of the equivalence relations $E_1, E_2, \dots, E_r$, the notation $\langle k, l\rangle$ means "the $k^{\text{th}}$ vertex of the left $K_n$ is equivalent to the $l^{\text{th}}$ vertex of the right $K_n$". Altogether, $E_j$ is defined as follows. We take the $j^{\text{th}}$ column of $A$ and the $j^{\text{th}}$ column of $B$, which are both permutations of $\{1,2,\dots,n\}$. Then, $E_j$ puts the $i^{\text{th}}$ entry of $A$'s column (a vertex in the left $K_n$) in the same $2$-element equivalence class as the $i^{\text{th}}$ entry of $B$'s column (a vertex in the right $K_n$), for each $i$.
The covering property of $A$ and $B$ tells us that $\bigcup_{j=1}^r E_j = E(G^c)$. If we take the $k^{\text{th}}$ vertex of the left $K_n$ and the $l^{\text{th}}$ vertex of the right $K_n$, then the pair $(k,l)$ must show up as $(a_{ij}, b_{ij})$ for some $i$ and $j$, which means that $E_j$ will put $k$ and $l$ in the same equivalence class.
Next, we check $\bigcap_{j=1}^r E_j = \Delta$. From the "column Latin" property, each column of $A$ and $B$ is a permutation of $\{1, \dots, n\}$, which means every $E_j$ matches each vertex of the left $K_n$ to exactly one vertex of the right $K_n$. Consider the $i^{\text{th}}$ vertex of the left $K_n$, for arbitrary $i$. Then by the covering property, some $E_j$ matches $i$ to $1$ (only), and some other $E_{j'}$ matches $i$ to $2$ (only). So $E_j \cap E_{j'}$ already does not match $i$ to anything: $\{i\}$ is a singleton equivalence class of $E_j \cap E_{j'}$. Therefore in $\bigcap_{j=1}^r E_j$, every vertex on the left (and by symmetry, every vertex on the right) is a singleton equivalence class.

As a side note: rather than "equivalence", we could equally well say "edge set of the disjoint union of complete graphs" (except for the minor detail that equivalence relations include the diagonal). This would give us a new phrasing of the definition of dimension: it's the least $n$ such that there are graphs $G_1, \dots, G_n$, each of which is the disjoint union of complete graphs, such that $G^c = G_1 \cup \dots \cup G_n$ and $E(G_1) \cap \dots \cap E(G_n) = \varnothing$. 
In this definition, the $E_1, \dots, E_r$ in the paper become perfect matchings between the left $K_n$ and the right $K_n$, which might be easier to visualize.
