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?