Hypothetical Syllogism/Formulation 5/Proof 2

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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

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

$\blacksquare$


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