Let be a directed graph with minimal strongly connected subgraphs and . Let and be directed edges with and .

We want to show that is not strongly connected.

Assume for a contradiction that is strongly connected. If and then is not simple, so we may assume otherwise.

As is strongly connected, there is a walk in . As only needs to occur once in this walk, we can assume that all the other edges used are in . That is is connected to and is connected to in .

If is connected to in then is strongly connected, contradicting that has no strongly connected subgraphs. Certainly there is a path from to in . But, it might contain . Is there a reason why it doesn't have to?

Um...if it does contain , then don't we need to replace with ...but that would mean we had a parallel arc, which is again a contradiction?

No. I'm not sure that makes sense. We're assuming is strongly connected for a contradiction (this gave us the path). The contradiction looks like it should be is strongly connected. But, we can't show is connected to in . If the path in from to contains , we can't use to fix it, as is not in (nor for that matter).