Hypothetical Syllogism/Formulation 3/Proof 1

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Theorem

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


Proof

Let us use the following abbreviations

\(\displaystyle \phi\) \(\text{ for }\) \(\displaystyle p \implies q\)
\(\displaystyle \psi\) \(\text{ for }\) \(\displaystyle q \implies r\)
\(\displaystyle \chi\) \(\text{ for }\) \(\displaystyle p \implies r\)


By the tableau method of natural deduction:

$\paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r} $
Line Pool Formula Rule Depends upon Notes
1 1 $\phi \land \psi$ Assumption (None)
2 1 $\phi$ Rule of Simplification: $\land \mathcal E_1$ 1
3 1 $\psi$ Rule of Simplification: $\land \mathcal E_2$ 1
4 1 $\chi$ Sequent Introduction 2, 3 Hypothetical Syllogism: Formulation 1
5 $\paren {\phi \land \psi} \implies \chi$ Rule of Implication: $\implies \mathcal I$ 1 – 4 Assumption 1 has been discharged

Expanding the abbreviations leads us back to:

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

$\blacksquare$


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