# Hypothetical Syllogism/Formulation 2/Proof 2

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## Theorem

\(\ds p\) | \(\implies\) | \(\ds q\) | ||||||||||||

\(\ds q\) | \(\implies\) | \(\ds r\) | ||||||||||||

\(\ds r\) | \(\) | \(\ds \) | ||||||||||||

\(\ds \vdash \ \ \) | \(\ds p\) | \(\) | \(\ds \) |

## 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): $\S 1.2$: Conditionals and Negation: Theorem $3$