Hypothetical Syllogism/Formulation 5

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Theorem

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


Proof 1

Let us use the following abbreviations

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


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: $\implies \II$ 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:

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

$\blacksquare$


Proof 2

This proof is derived in the context of the following proof system: Instance 2 of the Hilbert-style systems.

By the tableau method:

$\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$
Line Pool Formula Rule Depends upon Notes
1 $\paren {q \implies r} \implies \paren {\paren {p \lor q} \implies \paren {p \lor r} }$ Axiom $\text A 4$
2 $\paren {q \implies r} \implies \paren {\paren {\neg p \lor q} \implies \paren {\neg p \lor r} }$ Rule $\text {RST} 1$ 1 $\neg p \, / \, p$
3 $\paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ Rule $\text {RST} 2 \, (2)$ 2

$\blacksquare$


Sources

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