I have the following problem: let $G$ be a finite directed graph with $V$ vertices $v_i$ and $E$ (directed) edges $e_j$. I know that if an edge $e_k = (v_i, v_j)$ is in the graph, then the opposite edge $-e_k = (v_j, v_i)$ is not in the graph. I also know that the graph contains at least one cycle. My goal is to render the graph acyclic by swapping the direction of some edges **pertaining to at least one cycle**.

I figured this was simple induction reasoning, i.e. swap an edge in any given cycle thus removing it, and repeat. But I cannot find a proof that it is possible to pick an edge which would not create a new cycle when swapped, nor can I find a counter-example with a finite graph.

The lemma I'd like to prove would be: given a graph $G$ such as above, there exists edge $(v_i, v_j)$ in one of the cycles so that replacing it with its opposite $(v_j, v_i)$ decreases the number of cycles in $G$ by at least 1. However, I couldn't find anything related... Any idea?

Thanks for your help