Hypothetical Syllogism/Formulation 1/Proof 1
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Theorem
| \(\ds p\) | \(\implies\) | \(\ds q\) | ||||||||||||
| \(\ds q\) | \(\implies\) | \(\ds r\) | ||||||||||||
| \(\ds \vdash \ \ \) | \(\ds p\) | \(\implies\) | \(\ds r\) |
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 | $p$ | Assumption | (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 | ||
| 6 | 1, 2 | $p \implies r$ | Rule of Implication: $\implies \II$ | 3 – 5 | Assumption 3 has been discharged |
$\blacksquare$