Double Negation Elimination implies Law of Excluded Middle

Theorem
Let the Law of Double Negation Elimination be supposed to hold:
 * $\neg \neg p \vdash p$

Then the Law of the Excluded Middle likewise holds:
 * $\vdash p \lor \neg p$

Proof

 * align="right" | 3 ||
 * align="right" | 2
 * $p \lor \neg p$
 * $\lor \mathcal I_1$
 * 2
 * align="right" | 4 ||
 * align="right" | 1, 2
 * $\bot$
 * $\neg \mathcal E$
 * 3, 1
 * align="right" | 5 ||
 * align="right" | 1
 * $\neg p$
 * Proof by Contradiction
 * 2, 4
 * align="right" | 6 ||
 * align="right" | 1
 * $p \lor \neg p$
 * $\lor \mathcal I_2$
 * 5
 * align="right" | 7 ||
 * align="right" | 1
 * $\bot$
 * $\neg \mathcal E$
 * 6, 1
 * That also causes a contradiction
 * align="right" | 8 ||
 * align="right" |
 * $\neg \neg \left({p \lor \neg p}\right)$
 * $\neg \mathcal I$
 * 1, 7
 * align="right" | 9 ||
 * align="right" |
 * $p \lor \neg p$
 * $\neg \neg \mathcal E$
 * 8
 * }
 * align="right" |
 * $\neg \neg \left({p \lor \neg p}\right)$
 * $\neg \mathcal I$
 * 1, 7
 * align="right" | 9 ||
 * align="right" |
 * $p \lor \neg p$
 * $\neg \neg \mathcal E$
 * 8
 * }
 * 8
 * }
 * }

Comment
Thus the Law of Double Negation Elimination may be taken as an axiom instead of the Law of the Excluded Middle.