Rule of Exportation/Reverse Implication/Formulation 1/Proof

Theorem

 * $p \implies \left ({q \implies r}\right) \vdash \left ({p \land q}\right) \implies r$

Proof

 * align="right" | 3 ||
 * align="right" | 2
 * $p$
 * $\land \mathcal E_1$
 * 2
 * align="right" | 4 ||
 * align="right" | 2
 * $q$
 * $\land \mathcal E_2$
 * 2
 * align="right" | 5 ||
 * align="right" | 1, 2
 * $q \implies r$
 * $\implies \mathcal E$
 * 1, 3
 * align="right" | 6 ||
 * align="right" | 1, 2
 * $r$
 * $\implies \mathcal E$
 * 4, 5
 * align="right" | 7 ||
 * align="right" | 1
 * $\left ({p \land q}\right) \implies r$
 * $\implies \mathcal I$
 * 2 - 6
 * }
 * 4, 5
 * align="right" | 7 ||
 * align="right" | 1
 * $\left ({p \land q}\right) \implies r$
 * $\implies \mathcal I$
 * 2 - 6
 * }
 * 2 - 6
 * }
 * }