Biconditional Introduction/Sequent Form/Proof 1
Theorem
- $p \implies q, q \implies p \vdash p \iff q$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \implies q$ | Premise | (None) | ||
| 2 | 2 | $q \implies p$ | Premise | (None) | ||
| 3 | 1, 2 | $p \iff q$ | Biconditional Introduction: $\iff \II$ | 1, 2 |
$\blacksquare$