Hypothetical Syllogism/Formulation 5/Proof 1
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Theorem
- $\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$
Proof
Let us use the following abbreviations
| \(\ds \phi\) | \(\text{ for }\) | \(\ds p \implies q\) | ||||||||||||
| \(\ds \psi\) | \(\text{ for }\) | \(\ds q \implies r\) | ||||||||||||
| \(\ds \chi\) | \(\text{ for }\) | \(\ds p \implies r\) |
From Hypothetical Syllogism: Formulation 3 we have:
- $(1): \quad \vdash \paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r}$
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\psi \land \phi$ | Assumption | (None) | ||
| 2 | 1 | $\phi \land \psi$ | Sequent Introduction | 1 | Conjunction is Commutative | |
| 3 | 1 | $\chi$ | Sequent Introduction | 2 | Hypothetical Syllogism: Formulation 3 (see $(1)$ above) | |
| 4 | $\paren {\psi \land \phi} \implies \chi$ | Rule of Implication: $\implies \II$ | 1 – 3 | Assumption 1 has been discharged | ||
| 5 | $\psi \implies \paren {\phi \implies \chi}$ | Sequent Introduction | 4 | Rule of Exportation |
Expanding the abbreviations leads us back to:
- $\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$
$\blacksquare$