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:
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 |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation: Theorem $9$
- 2000: Michael R.A. Huth and Mark D. Ryan: Logic in Computer Science: Modelling and reasoning about systems ... (previous) ... (next): $\S 1.2.1$: Rules for natural deduction