# Hypothetical Syllogism/Formulation 2/Proof 1

Jump to navigation
Jump to search

## 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, 2 | $p \implies r$ | Sequent Introduction | 1, 2 | Hypothetical Syllogism: Formulation 1 | |

5 | 1, 2, 3 | $r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 4, 3 |

$\blacksquare$