Negation of Excluded Middle is False/Form 1
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Theorem
$\neg (p \lor \neg p) \vdash \bot$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\neg (p \lor \neg p)$ | Assumption | (None) | ||
| 2 | 1 | $\neg p \land \neg \neg p$ | Sequent Introduction | 1 | De Morgan's Laws | |
| 3 | 1 | $\neg p$ | Rule of Simplification: $\land \EE_1$ | 2 | ||
| 4 | 1 | $\neg\neg p$ | Rule of Simplification: $\land \EE_2$ | 2 | ||
| 5 | 1 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 3, 4 |
$\blacksquare$
Remark
The specific form of De Morgan's Laws used in this proof does not itself rely on Law of Excluded Middle in any way.