Proof by Contradiction/Variant 1

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Theorem

$\left({p \vdash \left({q \land \neg q}\right)}\right) \vdash \neg p$


Proof

By the tableau method of natural deduction:

$\left({p \vdash \left({q \land \neg q}\right)}\right) \vdash \neg p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \vdash \left({q \land \neg q}\right)$ Premise (None)
2 2 $p$ Assumption (None)
3 1, 2 $q \land \neg q$ Sequent Introduction 1, 2 by hypothesis
4 1, 2 $q$ Rule of Simplification: $\land \mathcal E_1$ 3
5 1, 2 $\neg q$ Rule of Simplification: $\land \mathcal E_2$ 3
6 1, 2 $\bot$ Principle of Non-Contradiction: $\neg \mathcal E$ 4, 5
7 1 $\neg p$ Proof by Contradiction: $\neg \mathcal I$ 2 – 6 Assumption 2 has been discharged


$\blacksquare$


Sources