Proof by Cases/Formulation 1/Reverse Implication

Theorem

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

Proof

 * align="right" | 3 ||
 * align="right" | 2
 * $p \lor q$
 * $\lor \mathcal I_1$
 * 2
 * align="right" | 4 ||
 * align="right" | 1, 2
 * $r$
 * $\implies \mathcal E$
 * 1, 3
 * align="right" | 5 ||
 * align="right" | 1
 * $p \implies r$
 * $\implies \mathcal I$
 * 2, 4
 * align="right" | 5 ||
 * align="right" | 1
 * $p \implies r$
 * $\implies \mathcal I$
 * 2, 4


 * align="right" | 7 ||
 * align="right" | 6
 * $p \lor q$
 * $\lor \mathcal I_2$
 * 6
 * align="right" | 8 ||
 * align="right" | 1, 6
 * $r$
 * $\implies \mathcal E$
 * 1, 7
 * align="right" | 9 ||
 * align="right" | 1
 * $q \implies r$
 * $\implies \mathcal I$
 * 6, 8
 * align="right" | 10 ||
 * align="right" | 1
 * $\left({p \implies r}\right) \land \left({q \implies r}\right)$
 * $\land \mathcal I$
 * 5, 9
 * }
 * align="right" | 10 ||
 * align="right" | 1
 * $\left({p \implies r}\right) \land \left({q \implies r}\right)$
 * $\land \mathcal I$
 * 5, 9
 * }
 * }
 * }