# Hypothetical Syllogism/Formulation 1/Proof 1

Jump to navigation
Jump to search

## Theorem

- $p \implies q, q \implies r \vdash p \implies r$

## 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: $\implies \mathcal I$ | 3 – 5 | Assumption 3 has been discharged |

$\blacksquare$