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

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


 * $\left({p \implies q}\right) \implies r \vdash \left({p \implies r}\right) \implies \left({q \implies r}\right)$

As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the LHS is $T$, that under the one on the RHS 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:
 * $\left({p \implies q}\right) \implies r = F$

but:
 * $\left({p \implies r}\right) \implies \left({q \implies r}\right) = T$

Hence the result:
 * $\left({p \implies q}\right) \implies r \vdash \left({p \implies r}\right) \implies \left({q \implies r}\right)$

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