Destructive Dilemma/Formulation 1/Proof 1
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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 | 3 | $\neg q \lor \neg s$ | Assumption | (None) | ||
| 4 | 4 | $\neg q$ | Assumption | (None) | ||
| 5 | 1, 4 | $\neg p$ | Modus Tollendo Tollens (MTT) | 1, 4 | ||
| 6 | 1, 4 | $\neg p \lor \neg r$ | Rule of Addition: $\lor \II_1$ | 5 | ||
| 7 | 7 | $\neg s$ | Assumption | (None) | ||
| 8 | 2, 7 | $\neg r$ | Modus Tollendo Tollens (MTT) | 2, 7 | ||
| 9 | 2, 7 | $\neg p \lor \neg r$ | Rule of Addition: $\lor \II_2$ | 8 | ||
| 10 | 1, 2, 3 | $\neg p \lor \neg r$ | Proof by Cases: $\text{PBC}$ | 3, 4 – 6, 7 – 9 | Assumptions 4 and 7 have been discharged | |
| 11 | 1, 2 | $\neg q \lor \neg s \implies \neg p \lor \neg r$ | Rule of Implication: $\implies \II$ | 3 – 10 | Assumption 3 has been discharged |
$\blacksquare$