Rule of Commutation/Conjunction/Formulation 1/Proof 1

Theorem

 * $p \land q \dashv \vdash q \land p$

Proof

 * align="right" | 2 ||
 * align="right" | 1
 * $p$
 * $\land \mathcal E_1$
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $q$
 * $\land \mathcal E_2$
 * 1
 * $q$
 * $\land \mathcal E_2$
 * 1


 * align="right" | 2 ||
 * align="right" | 1
 * $q$
 * $\land \mathcal E_1$
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $p$
 * $\land \mathcal E_2$
 * 1
 * $p$
 * $\land \mathcal E_2$
 * 1