Hypothetical Syllogism/Formulation 2/Proof 2
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Theorem
\(\ds p\) | \(\implies\) | \(\ds q\) | ||||||||||||
\(\ds q\) | \(\implies\) | \(\ds r\) | ||||||||||||
\(\ds p\) | \(\) | \(\ds \) | ||||||||||||
\(\ds \vdash \ \ \) | \(\ds r\) | \(\) | \(\ds \) |
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \implies q$ | Premise | (None) | ||
2 | 2 | $q \implies r$ | Premise | (None) | ||
3 | 3 | $p$ | Premise | (None) | ||
4 | 1, 3 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 3 | ||
5 | 1, 2, 3 | $r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 4 |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation: Theorem $3$