Say i have a hypothesis of the following form: $P \lor Q$ and a conclusion $\neg A$.
I try a proof by contradiction; so I assume $A$. Now what I am trying to do is break the hypothesis into cases, so:

* **Case 1:** I assume $P$ is true. This leads to a contradiction, so i conclude $\neg A$.
* **Case 2:** I assume $Q$ is true. This, however, does not lead to a contradiction.

In one of the cases I reached a contradiction and in the other I did not. Does this mean that the theorem is invalid as the conclusion is not reached even if the hypothesis is true (case 2) or is this a valid theorem?