Hypothetical Syllogism/Formulation 4/Proof 1

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Theorem

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


Proof

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:

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

$\blacksquare$