Conditional is not Right Self-Distributive/Formulation 1

Theorem
While this holds:
 * $\left({p \implies q}\right) \implies r \vdash \left({p \implies r}\right) \implies \left({q \implies r}\right)$

its converse does not:
 * $\left({p \implies r}\right) \implies \left({q \implies r}\right) \not \vdash \left({p \implies q}\right) \implies r$