Conditional is not Left Self-Distributive/Formulation 2

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Theorem

While this holds:

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

its converse does not:

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


Proof


Sources