True Statement is implied by Every Statement/Formulation 1/Proof 1
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Theorem
| \(\ds p\) | \(\) | \(\ds \) | ||||||||||||
| \(\ds \vdash \ \ \) | \(\ds q \implies p\) | \(\) | \(\ds \) |
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p$ | Premise | (None) | ||
| 2 | 1 | $\neg q \lor p$ | Rule of Addition: $\lor \II_2$ | 1 | ||
| 3 | 1 | $q \implies p$ | Sequent Introduction | 1 | Rule of Material Implication |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $2$: The Propositional Calculus $2$: $2$: Theorems and Derived Rules: Theorem $50$