Double Negation/Double Negation Elimination/Sequent Form/Formulation 2

Theorem

 * $\vdash \neg \neg p \implies p$

Proof

 * align="right" | 2 ||
 * align="right" | 1
 * $p$
 * $\neg \neg \mathcal E$
 * 1
 * using formulation 1 of this result
 * using formulation 1 of this result

Also see

 * Double Negation Elimination implies Law of Excluded Middle