Non-Equivalence of Proposition and Negation/Formulation 2
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Theorem
- $\vdash \neg \left({p \iff \neg p}\right)$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \iff \neg p$ | Assumption | (None) | ||
2 | 1 | $p \implies \neg p$ | Biconditional Elimination: $\iff \EE_1$ | 1 | ||
3 | 1 | $\neg p \implies p$ | Biconditional Elimination: $\iff \EE_2$ | 1 | ||
4 | 1 | $\bot$ | Sequent Introduction | 2,3 | Non-Equivalence of Proposition and Negation: Formulation 1 | |
5 | $\neg \left({p \iff \neg p}\right)$ | Proof by Contradiction: $\neg \II$ | 1 – 4 | Assumption 1 has been discharged |