# Non-Equivalence as Equivalence with Negation/Formulation 1

## Contents

## Theorem

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

This can be expressed as two separate theorems:

### Forward Implication

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

### Reverse Implication

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

## Proof 1

### Forward Implication: Proof

By the tableau method of natural deduction:

$\blacksquare$

#### Law of the Excluded Middle

This proof depends on the Law of the Excluded Middle.

This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom. This in turn invalidates this proof from an intuitionistic perspective.

### Reverse Implication: 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 |

## Proof 2

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.

$\begin{array}{|cccc||cccc|} \hline
\neg & (p & \iff & q) & (p & \iff & \neg & q) \\
\hline
F & F & T & F & T & F & T & F \\
T & F & F & T & T & T & F & T \\
T & T & F & F & F & T & T & F \\
F & T & T & T & F & F & F & T \\
\hline
\end{array}$

$\blacksquare$