# Factor Principles/Conjunction on Left/Formulation 1/Proof 1

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## Theorem

- $p \implies q \vdash \paren {r \land p} \implies \paren {r \land q}$

## Proof

By the tableau method of natural deduction:

Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|

1 | 1 | $p \implies q$ | Premise | (None) | ||

2 | $r \implies r$ | Law of Identity | (None) | This is a theorem so depends on nothing | ||

3 | 1 | $\paren {r \implies r} \land \paren {p \implies q}$ | Rule of Conjunction: $\land \mathcal I$ | 2, 1 | ||

4 | 1 | $\paren {r \land p} \implies \paren {r \land q}$ | Sequent Introduction | 3 | Praeclarum Theorema |

$\blacksquare$

## Sources

- 1965: E.J. Lemmon:
*Beginning Logic*... (previous) ... (next): $\S 1.3$: Conjunction and Disjunction: Theorem $18$