Proof by Cases/Formulation 2

Theorem
If both of two alternative cases imply the conclusion, then either one being true is enough to prove the conclusion true.
 * $\vdash \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \iff \left({\left({p \lor q}\right) \implies r}\right)$

This can be expressed as two separate theorems: