Rule of Commutation/Conjunction/Formulation 1
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Theorem
- $p \land q \dashv \vdash q \land p$
Proof 1
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \land q$ | Premise | (None) | ||
| 2 | 1 | $p$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
| 3 | 1 | $q$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
| 4 | 1 | $q \land p$ | Rule of Conjunction: $\land \II$ | 3, 2 |
$\Box$
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $q \land p$ | Premise | (None) | ||
| 2 | 1 | $q$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
| 2 | 1 | $p$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
| 4 | 1 | $p \land q$ | Rule of Conjunction: $\land \II$ | 3, 2 |
$\blacksquare$
Proof by Truth Table
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccc||ccc|} \hline p & \land & q & q & \land & p \\ \hline \F & \F & \F & \F & \F & \F \\ \F & \F & \T & \T & \F & \F \\ \T & \F & \F & \F & \F & \T \\ \T & \T & \T & \T & \T & \T \\ \hline \end{array}$
$\blacksquare$
Sources
- 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: Exercise $1.2: \ 2$
- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.3.3$