# Factor Principles/Conjunction on Left/Formulation 1

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## Theorem

- $p \implies q \vdash \left({r \land p}\right) \implies \left ({r \land q}\right)$

## Proof 1

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 |

$\blacksquare$

## Proof 2

By the tableau method of natural deduction:

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

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

2 | 2 | $r \land p$ | Assumption | (None) | ||

3 | 1, 2 | $p$ | Rule of Simplification: $\land \mathcal E_2$ | 2 | ||

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

5 | 1, 2 | $r$ | Rule of Simplification: $\land \mathcal E_1$ | 2 | ||

6 | 1, 2 | $r \land q$ | Rule of Conjunction: $\land \mathcal I$ | 5, 4 | ||

7 | 1 | $\paren {r \land p} \implies \paren {r \land q}$ | Rule of Implication: $\implies \mathcal I$ | 2 – 6 | Assumption 2 has been discharged |

$\blacksquare$

## Sources

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