Hypothetical Syllogism/Formulation 5/Proof 1

Theorem

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

Let us use the following abbreviations

$\ds \phi$

$\text{ for }$

$\ds p \implies q$

$\ds \psi$

$\text{ for }$

$\ds q \implies r$

$\ds \chi$

$\text{ for }$

$\ds p \implies r$

$(1): \quad \vdash \paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r}$

By the tableau method of natural deduction:

$\paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} } $
Line	Pool	Formula	Rule	Depends upon	Notes
1	1	$\psi \land \phi$	Assumption	(None)
2	1	$\phi \land \psi$	Sequent Introduction	1	Conjunction is Commutative
3	1	$\chi$	Sequent Introduction	2	Hypothetical Syllogism: Formulation 3 (see $(1)$ above)
4		$\paren {\psi \land \phi} \implies \chi$	Rule of Implication: $\implies \II$	1 – 3	Assumption 1 has been discharged
5		$\psi \implies \paren {\phi \implies \chi}$	Sequent Introduction	4	Rule of Exportation

Expanding the abbreviations leads us back to:

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

$\blacksquare$