Proof by Contradiction/Variant 2/Formulation 1

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Theorem

$p \implies q, p \implies \neg q \vdash \neg p$


Proof

By the tableau method of natural deduction:

$p \implies q, p \implies \neg q \vdash \neg p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies q$ Premise (None)
2 2 $p \implies \neg q$ Premise (None)
3 3 $p$ Assumption (None)
4 1, 3 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 3
5 2, 3 $\neg q$ Modus Ponendo Ponens: $\implies \mathcal E$ 2, 3
6 1, 2, 3 $\bot$ Principle of Non-Contradiction: $\neg \mathcal E$ 4, 5
7 1, 2 $\neg p$ Proof by Contradiction: $\neg \mathcal I$ 3 – 6 Assumption 3 has been discharged


$\blacksquare$


Sources