Hypothetical Syllogism/Formulation 4/Proof 1

Theorem

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

Proof
Let us use substitution instances as follows:

Using substitution instances leads us back to:
 * $\vdash \left({p \implies q}\right) \implies \left({\left({q \implies r}\right) \implies \left({p \implies r}\right)}\right)$