Destructive Dilemma/Formulation 1/Proof 2
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Theorem
\(\ds p \implies q\) | \(\) | \(\ds \) | ||||||||||||
\(\ds r \implies s\) | \(\) | \(\ds \) | ||||||||||||
\(\ds \vdash \ \ \) | \(\ds \neg q \lor \neg s \implies \neg p \lor \neg r\) | \(\) | \(\ds \) |
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \implies q$ | Premise | (None) | ||
2 | 2 | $r \implies s$ | Premise | (None) | ||
3 | 1, 2 | $\paren {p \land r} \implies \paren {q \land s}$ | Sequent Introduction | 1, 2 | Praeclarum Theorema | |
4 | 4 | $\neg q \lor \neg s$ | Assumption | (None) | ||
5 | 4 | $\neg \paren {q \land s}$ | Sequent Introduction | 4 | De Morgan's Laws: Disjunction of Negations | |
6 | 1, 2, 4 | $\neg \paren {p \land r}$ | Modus Tollendo Tollens (MTT) | 3, 5 | ||
7 | 1, 2 | $\neg p \lor \neg r$ | Sequent Introduction | 6 | De Morgan's Laws: Disjunction of Negations | |
8 | 1, 2 | $\neg q \lor \neg s \implies \neg p \lor \neg r$ | Rule of Implication: $\implies \II$ | 4 – 7 | Assumption 4 has been discharged |
$\blacksquare$