# 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 $42$