Modus Ponendo Ponens/Sequent Form/Proof 1
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Theorem
| \(\ds p\) | \(\implies\) | \(\ds q\) | ||||||||||||
| \(\ds p\) | \(\) | \(\ds \) | ||||||||||||
| \(\ds \vdash \ \ \) | \(\ds q\) | \(\) | \(\ds \) |
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \implies q$ | Premise | (None) | ||
| 2 | 2 | $p$ | Premise | (None) | ||
| 3 | 1, 2 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 2 |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation: Theorem $1$