# Biconditional is Reflexive/Proof 1

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## Theorem

- $\vdash p \iff p$

## Proof

By the tableau method of natural deduction:

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

1 | $p \implies p$ | Theorem Introduction | (None) | Law of Identity: Formulation 2 | ||

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

$\blacksquare$