Hypothetical Syllogism/Formulation 2/Proof 2

Theorem

$\ds p$

$\implies$

$\ds q$

$\implies$

$\ds r$

$\ds p$

$\ds $

$\ds \vdash \ \ $

$\ds r$

$\ds $

By the tableau method of natural deduction:

$p \implies q, q \implies r, p \vdash r$
Line	Pool	Formula	Rule	Depends upon
1	1	$p \implies q$	Premise	(None)
2	2	$q \implies r$	Premise	(None)
3	3	$p$	Premise	(None)
4	1, 3	$q$	Modus Ponendo Ponens: $\implies \mathcal E$	1, 3
5	1, 2, 3	$r$	Modus Ponendo Ponens: $\implies \mathcal E$	2, 4

$\blacksquare$

1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation: Theorem $3$