# Non-Equivalence as Equivalence with Negation/Formulation 1/Reverse Implication/Proof

Jump to navigation
Jump to search

## 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 |