Hypothetical Syllogism/Formulation 5

Theorem

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

Proof
Let us use substitution instances as follows:

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


 * align="right" | 2 ||
 * align="right" | 1
 * $\phi \land \psi$
 * Rule of Commutation
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $\chi$
 * $\textrm{HS}$
 * 2
 * Using formulation 1 of this result
 * 2
 * Using formulation 1 of this result


 * align="right" | 2 ||
 * align="right" |
 * $\psi \implies \left({\phi \implies \chi}\right)$
 * Rule of Exportation
 * 4
 * 4

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