Biconditional Equivalent to Biconditional of Negations/Formulation 1/Forward Implication
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Theorem
- $p \iff q \vdash \neg p \iff \neg q$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \iff q$ | Premise | (None) | ||
| 2 | 1 | $p \implies q$ | Biconditional Elimination: $\iff \EE_1$ | 1 | ||
| 3 | 1 | $\neg q \implies \neg p$ | Sequent Introduction | 2 | Rule of Transposition | |
| 4 | 1 | $q \implies p$ | Biconditional Elimination: $\iff \EE_2$ | 1 | ||
| 5 | 1 | $\neg p \implies \neg q$ | Sequent Introduction | 4 | Rule of Transposition | |
| 6 | 1 | $\neg p \iff \neg q$ | Biconditional Introduction: $\iff \II$ | 5, 3 |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $4$ The Biconditional: Exercise $1 \ \text{(c)}$