# Factor Principles/Disjunction on Left/Formulation 1

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## Theorem

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

## Proof 1

By the tableau method of natural deduction:

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

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

2 | $r \implies r$ | Theorem Introduction | (None) | Law of Identity: Formulation 2 | ||

3 | 1 | $\paren {r \lor p} \implies \paren {r \lor q}$ | Sequent Introduction | 2, 1 | Constructive Dilemma |

$\blacksquare$

## 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 | $r \lor p$ | Assumption | (None) | ||

3 | 3 | $r$ | Assumption | (None) | ||

4 | 3 | $r \lor q$ | Rule of Addition: $\lor \II_1$ | 3 | ||

5 | 5 | $p$ | Assumption | (None) | ||

6 | 1, 5 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 5 | ||

7 | 1, 5 | $r \lor q$ | Rule of Addition: $\lor \II_2$ | 6 | ||

8 | 1, 2 | $r \lor q$ | Proof by Cases: $\text{PBC}$ | 2, 3 – 4, 5 – 6 | Assumptions 3 and 5 have been discharged | |

9 | 1 | $\paren {r \lor p} \implies \paren {r \lor q}$ | Rule of Implication: $\implies \II$ | 2 – 7 | Assumption 2 has been discharged |

$\blacksquare$