# Hypothetical Syllogism/Formulation 4

## Theorem

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

## Proof 1

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:

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

$\blacksquare$

## Proof 2

This proof is derived in the context of the following proof system: instance 1 of a Hilbert proof system.

By the tableau method:

$\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$
Line Pool Formula Rule Depends upon Notes
1 1 $p$ Assumption (None)
2 2 $p \implies q$ Assumption (None)
3 1, 2 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 2
4 4 $q \implies r$ Assumption (None)
5 1, 2, 4 $r$ Modus Ponendo Ponens: $\implies \mathcal E$ 3, 4
6 2, 4 $p \implies r$ Deduction Rule 5
7 2 $\paren {q \implies r} \implies \paren {p \implies r}$ Deduction Rule 6
8 $\paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$ Deduction Rule 7

$\blacksquare$