# Factor Principles/Conjunction on Left/Formulation 1

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## Theorem

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

## 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: $\land \II$ | 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: $\land \EE_2$ | 2 | ||

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

5 | 1, 2 | $r$ | Rule of Simplification: $\land \EE_1$ | 2 | ||

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

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

$\blacksquare$

This article is complete as far as it goes, but it could do with expansion.In particular: Derive the this rule from from Factor Principles: Conjunction on Right by applying Conjunction is Commutative on the right hand side.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding this information.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Expand}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Sources

- 1965: E.J. Lemmon:
*Beginning Logic*... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction: Theorem $18$