Rule of Transposition/Formulation 1/Forward Implication/Proof

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Theorem

\(\ds p\) \(\implies\) \(\ds q\)
\(\ds \vdash \ \ \) \(\ds \neg q\) \(\implies\) \(\ds \neg p\)


Proof

By the tableau method of natural deduction:

$p \implies q \vdash \neg q \implies \neg p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies q$ Premise (None)
2 2 $\neg q$ Assumption (None)
3 1, 2 $\neg p$ Modus Tollendo Tollens (MTT) 1, 2
4 1 $\neg q \implies \neg p$ Rule of Implication: $\implies \II$ 2 – 3 Assumption 2 has been discharged

$\blacksquare$


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