Non-Equivalence as Equivalence with Negation/Formulation 1/Reverse Implication
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Theorem
- $\paren {p \iff \neg q} \vdash \neg \paren {p \iff q}$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \iff \neg q$ | Premise | (None) | ||
2 | 1 | $\neg \paren {p \iff \neg \neg q}$ | Sequent Introduction | 1 | Non-Equivalence as Equivalence with Negation: Forward Implication | |
3 | $\neg \neg q \iff q$ | Theorem Introduction | (None) | Double Negation | ||
4 | 1 | $\neg \paren {p \iff q}$ | Sequent Introduction | 2, 3 | Biconditional is Transitive |