# Conjunction with Negative Equivalent to Negation of Implication/Formulation 2

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## Theorem

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

This can be expressed as two separate theorems:

### Forward Implication

- $\vdash \paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} }$

### Reverse Implication

- $\vdash \paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q}$

## Proof 1

By the tableau method of natural deduction:

Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|

1 | 1 | $p \land \neg q$ | Assumption | (None) | ||

2 | 1 | $\neg \paren {p \implies q}$ | Sequent Introduction | 1 | Conjunction with Negative Equivalent to Negation of Implication: Formulation 1 | |

3 | $\paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} }$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged | ||

4 | 4 | $\neg \paren {p \implies q}$ | Assumption | (None) | ||

5 | 4 | $p \land \neg q$ | Sequent Introduction | 4 | Conjunction with Negative Equivalent to Negation of Implication: Formulation 1 | |

6 | $\paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q}$ | Rule of Implication: $\implies \II$ | 4 – 5 | Assumption 4 has been discharged | ||

7 | $\paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$ | Biconditional Introduction: $\iff \II$ | 3, 6 |

$\blacksquare$

## Proof by Truth Table

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.

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

$\blacksquare$