# Biconditional Elimination/Sequent Form

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## Contents

## Theorem

- $(1): \quad p \iff q \vdash p \implies q$
- $(2): \quad p \iff q \vdash q \implies p$

## Proof 1

### Form 1

By the tableau method of natural deduction:

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

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

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

$\blacksquare$

### Form 2

By the tableau method of natural deduction:

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

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

2 | 1 | $q \implies p$ | Biconditional Elimination: $\iff \mathcal E_2$ | 1 |

$\blacksquare$

## Proof 2

We apply the Method of Truth Tables.

$\begin{array}{|ccc||ccc|ccc|} \hline p & \iff & q & p & \implies & q & q & \implies & p \\ \hline F & T & F & F & T & F & F & T & F \\ F & F & T & F & T & T & T & F & F \\ T & F & F & T & F & F & F & T & T \\ T & T & T & T & F & T & T & T & T \\ \hline \end{array}$

As can be seen, when $p \iff q$ is true so are both $p \implies q$ and $q \implies p$.

$\blacksquare$

## Sources

- 2000: Michael R.A. Huth and Mark D. Ryan:
*Logic in Computer Science: Modelling and reasoning about systems*... (previous) ... (next): $\S 1.2.5$: An aside: proof by contradiction: Exercises $1.6: \ 7$