Hypothetical Syllogism/Formulation 1/Proof 2

Theorem

\(\ds p\)

\(\implies\)

\(\ds q\)

\(\ds q\)

\(\implies\)

\(\ds r\)

\(\ds \vdash \ \ \)

\(\ds p\)

\(\implies\)

\(\ds r\)

Proof

This proof uses $\mathscr H_2$, Instance 2 of the Hilbert proof systems.

Recall the sequent form of the Hypothetical Syllogism:

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

Applying the Rule of Detachment $\text {RST} 3$ twice, we obtain:

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

and subsequently:

$\vdash p \implies r$

as desired.

$\blacksquare$

Sources

• 1959: A.H. Basson and D.J. O'Connor: Introduction to Symbolic Logic (3rd ed.) ... (previous) ... (next): $\S 4.7$: The Derivation of Formulae: $DR \, 1$