Hypothetical Syllogism/Formulation 5

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Theorem

$\vdash \left({q \implies r}\right) \implies \left({\left({p \implies q}\right) \implies \left({p \implies r}\right)}\right)$


Proof 1

Let us use the following abbreviations

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


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:

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

$\blacksquare$


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:

$\vdash \left({q \implies r}\right) \implies \left({\left({p \implies q}\right) \implies \left({p \implies r}\right)}\right)$
Line Pool Formula Rule Depends upon Notes
1 $\left({q \implies r}\right) \implies \left({\left({p \lor q}\right) \implies \left ({p \lor r}\right)}\right)$ Axiom $A4$
2 $\left({q \implies r}\right) \implies \left({\left({\neg p \lor q}\right) \implies \left ({\neg p \lor r}\right)}\right)$ Rule $RST \, 1$ 1 $\neg p \, / \, p$
3 $\left({q \implies r}\right) \implies \left({\left({p \implies q}\right) \implies \left({p \implies r}\right)}\right)$ Rule $RST \, 2 \, (2)$ 2

$\blacksquare$


Sources