Hypothetical Syllogism/Formulation 4

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:

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

$\blacksquare$

This proof is derived in the context of the following proof system: instance 1 of a Hilbert proof system.

By the tableau method:

$\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$
Line	Pool	Formula	Rule	Depends upon
1	1	$p$	Assumption	(None)
2	2	$p \implies q$	Assumption	(None)
3	1, 2	$q$	Modus Ponendo Ponens: $\implies \mathcal E$	1, 2
4	4	$q \implies r$	Assumption	(None)
5	1, 2, 4	$r$	Modus Ponendo Ponens: $\implies \mathcal E$	3, 4
6	2, 4	$p \implies r$	Deduction Rule	5
7	2	$\paren {q \implies r} \implies \paren {p \implies r}$	Deduction Rule	6
8		$\paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$	Deduction Rule	7

$\blacksquare$

1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{I}$: 'NOT' and 'IF': $\S 5$: Theorem $\text{T4}$
1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(2) \ \text{(iii)}$