Proof by Contradiction/Variant 2/Formulation 1
Theorem
- $p \implies q, p \implies \neg q \vdash \neg p$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \implies q$ | Premise | (None) | ||
| 2 | 2 | $p \implies \neg q$ | Premise | (None) | ||
| 3 | 3 | $p$ | Assumption | (None) | ||
| 4 | 1, 3 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 3 | ||
| 5 | 2, 3 | $\neg q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 3 | ||
| 6 | 1, 2, 3 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 4, 5 | ||
| 7 | 1, 2 | $\neg p$ | Proof by Contradiction: $\neg \II$ | 3 – 6 | Assumption 3 has been discharged |
$\blacksquare$
Sources
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{I}$: 'NOT' and 'IF': $\S 3$: Exercises, Group $\text{I}: \ 18$
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction: Theorem $22$
- 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: Example $1.21$