Proof by Cases/Formulation 1/Forward Implication/Proof 1

Theorem

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

Proof

 * align="right" | 2 ||
 * align="right" | 1
 * $p \implies r$
 * $\land \mathcal E_1$
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $q \implies r$
 * $\land \mathcal E_2$
 * 1
 * align="right" | 4 ||
 * align="right" | 1
 * $p \lor q \implies r \lor r$
 * Sequent Introduction
 * 2, 3
 * Constructive Dilemma
 * align="right" | 1
 * $p \lor q \implies r \lor r$
 * Sequent Introduction
 * 2, 3
 * Constructive Dilemma


 * align="right" | 6 ||
 * align="right" | 1, 5
 * $r \lor r$
 * Modus Ponendo Ponens
 * 4, 5
 * align="right" | 7 ||
 * align="right" | 1, 5
 * $r$
 * Sequent Introduction
 * 6
 * Rule of Idempotence
 * align="right" | 8 ||
 * align="right" | 1
 * $p \lor q \implies r$
 * Rule of Implication
 * 5 - 7
 * }
 * $p \lor q \implies r$
 * Rule of Implication
 * 5 - 7
 * }
 * }