Proof by Contradiction/Variant 1
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Theorem
- $\paren {p \vdash \paren {q \land \neg q} } \vdash \neg p$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \vdash \paren {q \land \neg q}$ | 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: $\land \EE_1$ | 3 | ||
5 | 1, 2 | $\neg q$ | Rule of Simplification: $\land \EE_2$ | 3 | ||
6 | 1, 2 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 4, 5 | ||
7 | 1 | $\neg p$ | Proof by Contradiction: $\neg \II$ | 2 – 6 | Assumption 2 has been discharged |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction