Principle of Dilemma/Formulation 1/Forward Implication/Proof 3

Theorem

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

Proof
From the Constructive Dilemma we have:
 * $p \implies q, r \implies s \vdash p \lor r \implies q \lor s$

from which, changing the names of letters strategically:
 * $p \implies q, \neg p \implies q \vdash p \lor \neg p \implies q \lor q$

From the Law of Excluded Middle, we have:
 * $\vdash p \lor \neg p$

From the Rule of Idempotence we have:
 * $q \lor q \vdash q$

and the result follows by Hypothetical Syllogism.