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Thm 2.8:  An acyclic digraph, $D$ has a unique vertex basis consisting of all vertices with no incoming arcs.


Proof:  Let $B$ be the set of all vertices with no incoming arcs.  If $v \in B$.  Then since $v$ has 
no
incoming arcs, $v$ is only reachable from itself.  Thus $B$ must be a subset of any vertex base.  
Suppose $u \not\in B$.  Let $u_1 = u$.  Then $u_1$ has an incoming arc $(u_2, u_1)$.  If $u_2 \in B$,
then $u_1$ is reachable from a vertex in $B$.  If $u_2 \not\in B$ then $u_2$ has an incoming arc
$(u_3, u_2)$.

Suppose the path $u_n, ..., u_1$ is defined such that all vertices are distinct.  

If $u_n \in B$, then $u$ is reachable from a vertex in $B$.  

If $u_n \not\in B$ then $u_n$ has an incoming arc $(u_{n+1}, u_n)$.  If $u_{n+1} = u_i$ for some
$i = 1, ..., n$, then $u_{n+1}, u_n, ..., u_i$ is a cycle, a contradiction.  Hence all the vertices of
$u_{n+1}, u_n, ..., u_1$ are distinct.

Since the number of vertices of $D$ is finite, this process must eventually end, say with the path
$u_t, ..., u_1$.  Since we cannot continue this process, $u_t$ must not have any incoming arcs.  
Hence $u_t \in B$, and hence $u$ is reachable from a vertex in $B$.  Thus any vertex basis must be
contained in $B$.  Hence $B$ is the unique vertex basis of $D$.


\end