# Non-Equivalence as Equivalence with Negation/Formulation 2

## Theorem

$\vdash \neg \paren {p \iff q} \iff \paren {p \iff \neg q}$

## Proof

By the tableau method of natural deduction:

$\vdash \neg \paren {p \iff q} \iff \paren {p \iff \neg q}$
Line Pool Formula Rule Depends upon Notes
1 1 $\neg \paren {p \iff q}$ Assumption (None)
2 1 $p \iff \neg q$ Sequent Introduction 1 Non-Equivalence as Equivalence with Negation: Formulation 1
3 $\paren {\neg \paren {p \iff q} } \implies \paren {p \iff \neg q}$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged
4 4 $p \iff \neg q$ Assumption (None)
5 4 $\neg \paren {p \iff q}$ Sequent Introduction 4 Non-Equivalence as Equivalence with Negation: Formulation 1
6 $\paren {p \iff \neg q} \implies \paren {\neg \paren {p \iff q} }$ Rule of Implication: $\implies \II$ 4 – 5 Assumption 4 has been discharged
7 $\paren {\neg \paren {p \iff q} } \iff \paren {p \iff \neg q}$ Biconditional Introduction: $\iff \II$ 3, 6

$\blacksquare$