Double Negation/Double Negation Elimination/Sequent Form/Formulation 1/Proof

Theorem

 * $\neg \neg p \vdash p$

Proof

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


 * align="right" | 6 ||
 * align="right" | 1, 4
 * $p$
 * $\bot \mathcal E$
 * 5
 * 5

Also see

 * Double Negation Elimination implies Law of Excluded Middle