Hypothetical Syllogism/Formulation 4

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:

From Hypothetical Syllogism: Formulation 3 we have:
 * $\vdash \left({\left({p \implies q}\right) \land \left({q \implies r}\right)}\right) \implies \left({p \implies r}\right)$

By the tableau method of natural deduction:

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