Non-Equivalence of Proposition and Negation/Formulation 1
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Theorem
- $p \implies \neg p, \neg p \implies p \vdash \bot$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \implies \neg p$ | Premise | (None) | ||
2 | 2 | $\neg p \implies p$ | Premise | (None) | ||
3 | 1 | $\neg p$ | Sequent Introduction | 1 | Proof by Contradiction: Variant 3 | |
4 | 1, 2 | $p$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 3 | ||
5 | 1, 2 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 4, 3 |
$\blacksquare$