# Rule of Commutation/Disjunction/Formulation 2/Proof 1

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## Theorem

- $\vdash \paren {p \lor q} \iff \paren {q \lor p}$

## Proof

By the tableau method of natural deduction:

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

1 | 1 | $p \lor q$ | Assumption | (None) | ||

2 | 1 | $q \lor p$ | Sequent Introduction | 1 | Disjunction is Commutative | |

3 | $\left({p \lor q}\right) \implies \left({q \lor p}\right)$ | Rule of Implication: $\implies \mathcal I$ | 1 – 2 | Assumption 1 has been discharged | ||

4 | 4 | $q \lor p$ | Assumption | (None) | ||

5 | 4 | $p \lor q$ | Sequent Introduction | 4 | Disjunction is Commutative | |

6 | $\left({q \lor p}\right) \implies \left({p \lor q}\right)$ | Rule of Implication: $\implies \mathcal I$ | 4 – 5 | Assumption 4 has been discharged | ||

7 | $\left({p \lor q}\right) \iff \left({q \lor p}\right)$ | Biconditional Introduction: $\iff \mathcal I$ | 3, 6 |

$\blacksquare$