Biconditional with Tautology/Proof 1

Theorem

$p \iff \top \dashv \vdash p$


Proof

By the tableau method of natural deduction:

$p \iff \top \vdash p$
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:

$p \vdash p \iff \top$
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

$\blacksquare$

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