Hypothetical Syllogism/Formulation 1/Proof 3

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$p \implies q, q \implies r \vdash p \implies r$


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

Recall the sequent form of the Hypothetical Syllogism:

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

Applying the Rule of Detachment $RST3$ twice, we obtain:

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

and subsequently:

$\vdash p \implies r$

as desired.