Rule of Transposition/Formulation 2/Forward Implication/Proof
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Theorem
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- $\vdash \paren {p \implies q} \implies \paren {\neg q \implies \neg p}$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \implies q$ | Assumption | (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 | |
| 5 | $\paren {p \implies q} \implies \paren {\neg q \implies \neg p}$ | Rule of Implication: $\implies \II$ | 1 – 4 | Assumption 1 has been discharged |
$\blacksquare$
Sources
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{I}$: 'NOT' and 'IF': $\S 5$: Theorem $\text{T13}$
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $2$: The Propositional Calculus $2$: $2$: Theorems and Derived Rules: Theorem $42$