# Implication is Left Distributive over Disjunction/Formulation 1/Forward Implication

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## Theorem

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

## Proof

By the tableau method of natural deduction:

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

1 | 1 | $p \implies \left({q \lor r}\right)$ | Assumption | (None) | ||

2 | 2 | $p$ | Assumption | (None) | ||

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

4 | 2 | $p$ | Law of Identity | 2 | ||

5 | 5 | $q$ | Assumption | (None) | ||

6 | 5 | $p \implies q$ | Rule of Implication: $\implies \mathcal I$ | 4 – 5 | Assumption 4 has been discharged | |

7 | 5 | $\left({p \implies q}\right) \lor \left({p \implies r}\right)$ | Rule of Addition: $\lor \mathcal I_1$ | 6 | ||

8 | 8 | $r$ | Assumption | (None) | ||

9 | 8 | $p \implies r$ | Sequent Introduction | 8 | True Statement is implied by Every Statement | |

10 | 8 | $\left({p \implies q}\right) \lor \left({p \implies r}\right)$ | Rule of Addition: $\lor \mathcal I_2$ | 9 | ||

11 | 1 | $\left({p \implies q}\right) \lor \left({p \implies r}\right)$ | Proof by Cases: $\text{PBC}$ | 3, 2 – 7, 8 – 10 | Assumptions 2 and 8 have been discharged |

$\blacksquare$