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. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.