Principle of Dilemma/Formulation 1/Forward Implication/Proof 2

Theorem

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

Proof

 * align="right" | 4 ||
 * align="right" |
 * $p \lor \neg p$
 * Law of Excluded Middle
 * (None)
 * (None)


 * align="right" | 6 ||
 * align="right" | 1, 5
 * $q$
 * $\implies \mathcal E$
 * 5, 2
 * 5, 2


 * align="right" | 8 ||
 * align="right" | 1, 7
 * $q$
 * $\implies \mathcal E$
 * 7, 3
 * align="right" | 9 ||
 * align="right" | 1
 * $q$
 * $\lor \mathcal E$
 * 4, 5-6, 7-8
 * $q$
 * $\lor \mathcal E$
 * 4, 5-6, 7-8