Clavius's Law/Formulation 1

Theorem
If, from the negation of a proposition $p$ we can derive $p$ itself, we may conclude $p$:
 * $\neg p \implies p \vdash p$

Proof

 * align="right" | 2 ||
 * align="right" |
 * $p \lor \neg p$
 * LEM
 * (None)
 * (None)


 * align="right" | 4 ||
 * align="right" | 1, 3
 * $p \,$
 * $\implies \mathcal E$
 * 1, 3
 * 1, 3