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:
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:
- $\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$
$\blacksquare$