Hypothetical Syllogism/Formulation 3

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Theorem

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


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 {\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: $\land \EE_1$ 1
3 1 $\psi$ Rule of Simplification: $\land \EE_2$ 1
4 1 $\chi$ Sequent Introduction 2, 3 Hypothetical Syllogism: Formulation 1
5 $\paren {\phi \land \psi} \implies \chi$ Rule of Implication: $\implies \II$ 1 – 4 Assumption 1 has been discharged

Expanding the abbreviations leads us back to:

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

$\blacksquare$


Proof by Truth Table

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.

$\begin{array}{|ccccccc|c|ccc|} \hline

((p & \implies & q) & \land & (q & \implies & r)) & \implies & (p & \implies & r) \\ \hline \F & \T & \F & \T & \F & \T & \F & \T & \F & \T & \F \\ \F & \T & \F & \T & \F & \T & \T & \T & \F & \T & \T \\ \F & \T & \T & \T & \T & \F & \F & \T & \F & \T & \F \\ \F & \T & \T & \T & \T & \T & \T & \T & \F & \T & \T \\ \T & \F & \F & \F & \F & \T & \F & \T & \T & \F & \F \\ \T & \F & \F & \T & \F & \T & \T & \T & \T & \T & \T \\ \T & \T & \T & \F & \T & \F & \F & \T & \T & \F & \F \\ \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$

$\blacksquare$


Sources

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