Natural deduction proof that $(P\leftrightarrow \neg P)$ is a contradiction, without first deriving $(P\vee \neg P)$ I'm looking to prove that $(P \leftrightarrow \neg P)$ is a contradiction using a natural deduction proof (which is to say, I want a proof to show $(P\leftrightarrow \neg P)\vdash Q$). In case it helps, the specific system I'm working in is as outlined in Halbach's Logic Manual (tree structure, introduction and elimination rules for each connective; see the link below), but it's the overall structure of the proof I'm struggling with.
Given a proof that shows $\vdash (P \vee \neg P)$ I can transform this into the desired proof, but that generates a very large tree given the simplicity of the sentence, because the proof for $\vdash (P \vee \neg P)$ is fairly long itself.
I can't shake the feeling there must be a more straightforward (even if still indirect) proof, but haven't been able to find it so far.

Edit: As found by lemontree, the ruleset I am using is listed here.

(Posted after answer accepted.) Using a form of natural deduction:


