Double Negation/Double Negation Introduction/Sequent Form/Formulation 1/Proof
Jump to navigation
Jump to search
Theorem
- $p \vdash \neg \neg p$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p$ | Premise | (None) | ||
2 | 2 | $\neg p$ | Assumption | (None) | ||
3 | 1, 2 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 1, 2 | ||
4 | 1 | $\neg \neg p$ | Proof by Contradiction: $\neg \II$ | 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
- 2000: Michael R.A. Huth and Mark D. Ryan: Logic in Computer Science: Modelling and reasoning about systems ... (previous) ... (next): $\S 1.2.2$: Derived rules