If no circuit then unique vertex basis Show the vertex basis in a directed graph is unique if there is no sequence of directed edges that forms a circuit in the graph.
Is this right? Or I should provide more explanation
Let A and B be two different vertex basis of graph G. Since A is vertex basis, then there exists some directed path that takes a vertex a' in A to a vertex b' in B. Similarly, there is a directed path taking vertex b' to a'. Contradiction since this creates a circuit a'-x1-x2....-xn-b'y1-y2-.....yn-a'
 A: To begin, the definition of a vertex basis is:

A vertex basis is a set of vertices where there’s a path to every vertex outside this set from vertices of this set, and there’s no path from any vertex in the set to another vertex in the set.  (Source)

Regarding the proposed proof, this is not necessarily true:

Similarly, there is a directed path taking vertex $b'$ to $a'$

since $a'$ and $b'$ have been determined.  All we know is that there is a path from some vertex $v \in B$ to $a'$ (and maybe $v \neq b'$).
To prove the claim:


*

*We conjecture that the unique vertex basis will be the set of all vertices of in-degree $0$.  It immediately follows that this is a subset of any vertex basis.

*Suppose a vertex $v$ of in-degree $\geq 1$ is in a hypothetical vertex basis.  We walk backwards (i.e., against the edge direction) until either (a) we reach a vertex with in-degree $0$ (which is in the vertex basis, giving a contradiction) or (b) we reach a vertex we've seen before (which gives rise to a circuit).
