Conditional is not Right Self-Distributive/Formulation 1

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

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

Proof
We apply the Method of Truth Tables to the proposition:


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

As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the is $\T$, that under the one on the  is also $\T$:

$\begin{array}{|ccccc||ccccccc|} \hline (p & \implies & q) & \implies & r & (p & \implies & r) & \implies & (q & \implies & r) \\ \hline \F & \T & \F & \F & \F & \F & \T & \F & \T & \F & \T & \F \\ \F & \T & \F & \T & \T & \F & \T & \T & \T & \F & \T & \T \\ \F & \T & \T & \F & \F & \F & \T & \F & \F & \T & \F & \F \\ \F & \T & \T & \T & \T & \F & \T & \T & \T & \T & \T & \T \\ \T & \F & \F & \T & \F & \T & \F & \F & \T & \F & \F & \F \\ \T & \F & \F & \T & \T & \T & \T & \T & \T & \F & \T & \T \\ \T & \T & \T & \F & \F & \T & \F & \F & \T & \T & \F & \F \\ \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$

The two formulas are not equivalent, as the relevant columns do not match exactly.

For example, when $p = q = r = \F$ we have that:
 * $\paren {p \implies q} \implies r = \F$

but:
 * $\paren {p \implies r} \implies \paren {q \implies r} = \T$

Hence the result:
 * $\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$

but:
 * $\paren {p \implies r} \implies \paren {q \implies r} \not \vdash \paren {p \implies q} \implies r$