Conditional is not Right Self-Distributive/Formulation 2

Theorem
While this holds:
 * $\vdash \paren {\paren {p \implies q} \implies r} \implies \paren {\paren {p \implies r} \implies \paren {q \implies r} }$

its converse does not:
 * $\not \vdash \paren {\paren {p \implies r} \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies r}$