Conditional is not Right Self-Distributive

Theorem

Formulation 1

While this holds:

$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$

its converse does not:

$\paren {p \implies r} \implies \paren {q \implies r} \not \vdash \paren {p \implies q} \implies r$

Formulation 2

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:

$\not \vdash \paren {\paren {p \implies r} \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies r}$