Biconditional Introduction/Sequent Form/Proof 1
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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 |
$\blacksquare$