Hypothetical Syllogism/Formulation 3

Theorem

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

Proof
Let us use substitution instances as follows:


 * align="right" | 2 ||
 * align="right" | 1
 * $\phi$
 * $\land \mathcal E_1$
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $\psi$
 * $\land \mathcal E_1$
 * 1
 * align="right" | 4 ||
 * align="right" | 1
 * $\chi$
 * $\textrm{HS}$
 * 2, 3
 * Using formulation 1 of this result
 * align="right" | 5 ||
 * align="right" |
 * $\left({\phi \land \psi}\right) \implies \chi$
 * $\implies \mathcal I$
 * 1, 4
 * }
 * align="right" | 5 ||
 * align="right" |
 * $\left({\phi \land \psi}\right) \implies \chi$
 * $\implies \mathcal I$
 * 1, 4
 * }
 * }

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