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:

$p \land q \vdash q \land p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \land q$ Premise (None)
2 1 $p$ Rule of Simplification: $\land \mathcal E_1$ 1
3 1 $q$ Rule of Simplification: $\land \mathcal E_2$ 1
4 1 $q \land p$ Rule of Conjunction: $\land \mathcal I$ 3, 2

$\Box$


By the tableau method of natural deduction:

$q \land p \vdash p \land q$
Line Pool Formula Rule Depends upon Notes
1 1 $q \land p$ Premise (None)
2 1 $q$ Rule of Simplification: $\land \mathcal E_1$ 1
2 1 $p$ Rule of Simplification: $\land \mathcal E_2$ 1
4 1 $p \land q$ Rule of Conjunction: $\land \mathcal I$ 3, 2

$\blacksquare$


Proof 2

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.


$\begin{array}{|ccc||ccc|} \hline p & \land & q & q & \land & p \\ \hline F & F & F & F & F & F \\ F & F & T & T & F & F \\ T & F & F & F & F & T \\ T & T & T & T & T & T \\ \hline \end{array}$

$\blacksquare$


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