# Biconditional with Contradiction

Jump to navigation
Jump to search

## Theorem

A biconditional with a contradiction:

- $p \iff \bot \dashv \vdash \neg p$

## Proof by Natural Deduction

By the tableau method of natural deduction:

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

1 | 1 | $p \iff \bot$ | Premise | (None) | ||

2 | 1 | $p \implies \bot$ | Biconditional Elimination: $\iff \mathcal E_1$ | 1 | ||

3 | 1 | $\neg p$ | Sequent Introduction | 2 | Contradictory Consequent |

$\Box$

By the tableau method of natural deduction:

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

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

2 | 1 | $p \implies \bot$ | Sequent Introduction | 1 | Contradictory Consequent | |

3 | $\top$ | Rule of Top-Introduction: $\top \mathcal I$ | (None) | |||

4 | $\bot \implies p$ | Sequent Introduction | 3 | Contradictory Antecedent | ||

5 | 1 | $p \iff \bot$ | Biconditional Introduction: $\iff \mathcal I$ | 2, 4 |

$\blacksquare$

## Proof by Truth Table

We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.

$\begin{array}{|c|ccc||cc|} \hline p & p & \iff & \bot & \neg & p \\ \hline F & F & T & F & T & F \\ T & T & F & F & F & T \\ \hline \end{array}$

$\blacksquare$

## Sources

- 2012: M. Ben-Ari:
*Mathematical Logic for Computer Science*(3rd ed.) ... (previous) ... (next): $\S 2.3.3$