Not only do I not understand how to do this, but I don't comprehend the solution:

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Here, supposons means assume, and donc means thus.

I'm specifically confused with line 5, for which I don't understand the rule $\implies E 2,3$ in the slightest; the lines it is referring to are not even implications! Any help would be appreciated, thanks!

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    $\begingroup$ $\neg P$ is often defined as $P\Rightarrow\bot$ which is apparently what's going here. $\endgroup$ – Derek Elkins May 16 at 19:17
  • $\begingroup$ @DerekElkins Ah of course, I'd totally forgotten about that. But in that case, how do we even reach bottom from $P(x)$? Thanks. $\endgroup$ – iaskdumbstuff May 16 at 19:20
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    $\begingroup$ Line 5 cites $\to E 2,4$, not $\to E 2,3$ (as you wrote). So $\to E$ is done on l. 2 $\neg (P(x) \lor R(x)) \equiv (P(x) \lor R(x)) \to \bot$ and l.4 $(P(x) \lor R(x))$ to derive $\bot$. Does this resolve your confusion? $\endgroup$ – lemontree May 16 at 19:52
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    $\begingroup$ As for the rest, you'd have to a bit more specific about what else you'd like to have explained. $\endgroup$ – lemontree May 16 at 19:53
  • $\begingroup$ @lemontree Of course! I'm so dumb; the answer to both of my questions was that a negation can be described as an implication of bottom. That clears everything for me, the rest I understand. Thanks very much to the both of you! $\endgroup$ – iaskdumbstuff May 16 at 20:13

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