Modus Ponendo Tollens/Sequent Form/Case 1
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Theorem
- $\neg \paren {p \land q}, p \vdash \neg q$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\neg \paren {p \land q}$ | Premise | (None) | ||
2 | 2 | $p$ | Premise | (None) | ||
3 | 3 | $q$ | Assumption | (None) | ||
4 | 2, 3 | $p \land q$ | Rule of Conjunction: $\land \II$ | 2, 3 | ||
5 | 1, 2, 3 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 4, 1 | ||
6 | 1, 2 | $\neg q$ | Proof by Contradiction: $\neg \II$ | 3 – 5 | Assumption 3 has been discharged |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Ruless: Theorem $34$