Hypothetical Syllogism/Formulation 1/Proof 2
Jump to navigation
Jump to search
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: $DR \, 1$