Hypothetical Syllogism/Formulation 5/Proof 2

Theorem

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


Proof

This proof is derived in the context of the following proof system: Instance 2 of the Hilbert-style systems.

By the tableau method:

$\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$
Line Pool Formula Rule Depends upon Notes
1 $\paren {q \implies r} \implies \paren {\paren {p \lor q} \implies \paren {p \lor r} }$ Axiom $\text A 4$
2 $\paren {q \implies r} \implies \paren {\paren {\neg p \lor q} \implies \paren {\neg p \lor r} }$ Rule $\text {RST} 1$ 1 $\neg p \, / \, p$
3 $\paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ Rule $\text {RST} 2 \, (2)$ 2

$\blacksquare$


Sources

  • 1959: A.H. Basson and D.J. O'Connor: Introduction to Symbolic Logic (3rd ed.) ... (previous) ... (next): $\S 4.3$: Derivable Formulae: Example $2$
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