# 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:

$\paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$
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: $\implies \mathcal I$ 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:

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

$\blacksquare$