Non-Equivalence of Proposition and Negation/Formulation 1

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Theorem

$p \implies \neg p, \neg p \implies p \vdash \bot$


Proof

By the tableau method of natural deduction:

$p \implies \neg p, \neg p \implies p \vdash \bot$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies \neg p$ Premise (None)
2 2 $\neg p \implies p$ Premise (None)
3 1 $\neg p$ Sequent Introduction 1 Proof by Contradiction: Variant 3
4 1, 2 $p$ Modus Ponendo Ponens: $\implies \mathcal E$ 2, 3
5 1, 2 $\bot$ Principle of Non-Contradiction: $\neg \EE$ 4, 3

$\blacksquare$