Hypothetical Syllogism/Formulation 2/Proof 2

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Theorem

\(\ds p\) \(\implies\) \(\ds q\)
\(\ds q\) \(\implies\) \(\ds r\)
\(\ds r\) \(\) \(\ds \)
\(\ds \vdash \ \ \) \(\ds p\) \(\) \(\ds \)


Proof

By the tableau method of natural deduction:

$p \implies q, q \implies r, p \vdash r$
Line Pool Formula Rule Depends upon Notes
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

$\blacksquare$


Sources