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

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## Theorem

- $p \lor q \dashv \vdash q \lor p$

## Proof

By the tableau method of natural deduction:

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

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

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

3 | 2 | $q \lor p$ | Rule of Addition: $\lor \II_2$ | 2 | ||

4 | 4 | $p$ | Assumption | (None) | ||

5 | 4 | $q \lor p$ | Rule of Addition: $\lor \II_1$ | 4 | ||

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

$\blacksquare$

By the tableau method of natural deduction:

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

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

2 | 2 | $q$ | Assumption | (None) | ||

3 | 2 | $p \lor q$ | Rule of Addition: $\lor \II_2$ | 2 | ||

4 | 4 | $p$ | Assumption | (None) | ||

5 | 4 | $p \lor q$ | Rule of Addition: $\lor \II_1$ | 4 | ||

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

$\blacksquare$

## Sources

- 1965: E.J. Lemmon:
*Beginning Logic*... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction: Theorem $19$ - 2000: Michael R.A. Huth and Mark D. Ryan:
*Logic in Computer Science: Modelling and reasoning about systems*... (previous) ... (next): $\S 1.2.1$: Rules for natural deduction