Biconditional with Tautology/Proof 1
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Theorem
- $p \iff \top \dashv \vdash p$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \iff \top$ | Premise | (None) | ||
| 2 | $\top$ | Rule of Top-Introduction: $\top \II$ | (None) | |||
| 3 | 1 | $\top \implies p$ | Biconditional Elimination: $\iff \EE_2$ | 1 | ||
| 4 | 1 | $p$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 3 |
$\Box$
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\top$ | Premise | (None) | ||
| 2 | 2 | $p$ | Assumption | (None) | ||
| 3 | $\top$ | Rule of Top-Introduction: $\top \II$ | (None) | |||
| 4 | $p \implies \top$ | Rule of Implication: $\implies \II$ | 2 – 3 | Assumption 2 has been discharged | ||
| 5 | 2 | $\top \implies p$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged | |
| 6 | 2 | $p \iff \top$ | Biconditional Introduction: $\iff \II$ | 4, 5 |
$\blacksquare$