Anti-symmetric if $AB= 1$ and $BA=0$ but every vertex has loops? I'm creating a directed graph from an adjacency list. The $0$ present that there is no relation while the $1$ represent that there is.
So i have a quick question regarding this.
Lets assume that $AB = 1$ that is that it has a connection.
$BA = 0$ which means that is does not have a connection. This continues throughout the graph. So we assume that it is anti symmetric. However the graph has loops. So $BB = 1$  $CC= 1$ etc.
It still considered anti symmetric?

Yes. A relation $\mathrel{R}$ is antisymmetric if 
$$x\mathrel{R}y\quad\text{and}\quad y\mathrel{R}x\quad\text{implies}\quad x=y\;.$$
This means that if $x\ne y$, you can’t have both $x\mathrel{R}y$ and $y\mathrel{R}x$: you can have at most one of them. It says nothing at all about what you can (or must) have when $x=y$.
In terms of the associated graph, if you never have both an edge $A\to B$ and an edge $B\to A$ when $A\ne B$, then the relation is antisymmetric; you can have as many or as few loops as you like. (If you have a loop at every vertex, the relation is reflexive.)
