Rule of Commutation/Disjunction/Formulation 1
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Theorem
- $p \lor q \dashv \vdash q \lor p$
Proof 1
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 |
$\blacksquare$
Proof 2
We apply the Method of Truth Tables.
As can be seen by inspection, in both cases, the truth values under the main connectives match for all boolean interpretations.
- $\begin{array}{|ccc||ccc|} \hline p & \lor & q & q & \lor & p \\ \hline \F & \F & \F & \F & \F & \F \\ \F & \T & \T & \T & \T & \F \\ \T & \T & \F & \F & \T & \T \\ \T & \T & \T & \T & \T & \T \\ \hline \end{array}$
$\blacksquare$
Also see
Sources
- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.3.3$