Hypothetical Syllogism/Formulation 4/Proof 1

Theorem

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

Let us use the following abbreviations

$\ds \phi$

$\text{ for }$

$\ds p \implies q$

$\ds \psi$

$\text{ for }$

$\ds q \implies r$

$\ds \chi$

$\text{ for }$

$\ds p \implies r$

By the tableau method of natural deduction:

$\paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} } $
Line		Pool	Formula	Rule	Depends upon	Notes
1			$\paren {\phi \land \psi} \implies \chi$	Theorem Introduction	(None)	Hypothetical Syllogism: Formulation 3
2			$\phi \implies \paren {\psi \implies \chi}$	Sequent Introduction	1	Rule of Exportation

Expanding the abbreviations leads us back to:

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

$\blacksquare$