Biconditional with Contradiction/Proof 1

Theorem

$p \iff \bot \dashv \vdash \neg p$


Proof

By the tableau method of natural deduction:

$p \iff \bot \vdash \neg p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \iff \bot$ Premise (None)
2 1 $p \implies \bot$ Biconditional Elimination: $\iff \EE_1$ 1
3 1 $\neg p$ Sequent Introduction 2 Contradictory Consequent

$\Box$


By the tableau method of natural deduction:

$\neg p \vdash p \iff \bot$
Line Pool Formula Rule Depends upon Notes
1 1 $\neg p$ Assumption (None)
2 1 $p \implies \bot$ Sequent Introduction 1 Contradictory Consequent
3 $\top$ Rule of Top-Introduction: $\top \II$ (None)
4 $\bot \implies p$ Sequent Introduction 3 Contradictory Antecedent
5 1 $p \iff \bot$ Biconditional Introduction: $\iff \II$ 2, 4

$\blacksquare$

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