Hypothetical Syllogism/Formulation 1/Proof 2

Theorem

$\ds p$

$\implies$

$\ds q$

$\implies$

$\ds r$

$\ds \vdash \ \ $

$\ds p$

$\implies$

$\ds r$

This proof uses $\mathscr H_2$, Instance 2 of the Hilbert proof systems.

$\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$

Applying the Rule of Detachment $\text {RST} 3$ twice, we obtain:

$\vdash \paren {p \implies q} \implies \paren {p \implies r}$

and subsequently:

$\vdash p \implies r$

as desired.

$\blacksquare$

1959: A.H. Basson and D.J. O'Connor: Introduction to Symbolic Logic (3rd ed.) ... (previous) ... (next): $\S 4.7$: The Derivation of Formulae: $DR \, 1$