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:

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


Proof


This theorem requires a proof.
In particular: Can be seen to be logically equivalent to Conditional is not Right Self-Distributive/Formulation 1 by application of the Rule of Implication and Modus Ponendo Ponens.
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