Hypothetical Syllogism/Formulation 3/Proof 1

Theorem

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

Proof
Let us use the following abbreviations

Expanding the abbreviations leads us back to:
 * $\paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r}$